Selina Concise Mathsclass 6 ICSE Solutions

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations

August 11, 2022June 1, 2018 by Prasanna

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations (Including Word Problems)

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations (Including Word Problems)

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 6 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 6 with Free PDF download option. Selina Publishers Concise Mathematics for Class 6 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

Selina Class 6 Maths ICSE Solutions Physics Chemistry Biology Geography History & Civics

IMPORTANT POINTS

1. Simple Equations : A mathematical statement, which states that two expressions are equal, is called simple equation.
2. Properties of Simple Equation :
(i) If same quantity is added to both the sides of simple equation, the sums are equal.
For Example :
x = 6 ⇒ x + a = 6 + a [Adding a on both the sides]
(ii) If same quantity is subtracted from both the sides of simple equation, the remainders are equal.
For Example :
x = 6 ⇒ x-a = 6 – a [Subtracting a on both the sides]
(iii) If both the sides of an equation are multiplied by the same quantity, the products are equal.
For Example :
x = 6 ⇒ a x x = a x 6 i.e. ax = 6a [Multiplying both the sides by a]
(iv) If both the sides of simple equation are divided by the same quantity, the quotients are equal.
For Example :
x = 6 ⇒ \(\frac { x }{ a }\) = \(\frac { 6 }{ a }\) [Dividing both the sides by a]

Simple (Linear) Equations Exercise 22A – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Solve:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 1

Solution:

Question 2.
Solve:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 7
Solution:

Question 3.
Solve:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 11
Solution:

Question 4.
Solve:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 13
Solution:

Question 5.
Solve:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 16
Solution:

Simple (Linear) Equations Exercise 22B – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Solve:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 20

Solution:

Question 2.
Solve:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 25
Solution:

Question 3.
Solve:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 29
Solution:

Simple (Linear) Equations Exercise 22C – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
5 – x = 3
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 33

Question 2.
2 – y = 8
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 34

Question 3.
8.4 – x = -2
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 35

Question 4.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 36
Solution:

Question 5.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 38
Solution:

Question 6.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 40
Solution:

Question 7.
1.6z = 8
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 156
Question 8.
3a = – 2.1
Solution:

Question 9.

Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 45

Question 10.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 46
Solution:

Question 11.
– 5x = 10
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 48

Question 12.
2.4z = -4.8
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 49

Question 13.
2y – 5 = -11
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 50

Question 14.
2x + 4.6 = 8
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 51

Question 15.
5y – 3.5 = 10
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 52

Question 16.
3x + 2 = -2.2
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 53

Question 17.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 54
Solution:

Question 18.

Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 57

Question 19.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 58
Solution:

Question 20.
-3y – 2 = 10
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 60

Question 21.
4z – 5 = 3 – z
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 61

Question 22.
7x -3x +2 =22
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 62

Question 23.
6y + 3 = 2y + 11
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 63

Question 24.
3 (x+5) = 18
Solution:
3 (x+5) = 18
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 64

Question 25.
5 (x-2) -2 (x+2) = 3
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 65

Question 26.
(5x-3) 4=3
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 66

Question 27.
3(2x+1) -2(x-5) -5 (5-2x) = 16
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 67

Simple (Linear) Equations Exercise 22D – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
A number increased by 17 becomes 54. Find the number.
Solution:
Let the required number = x
∴ According to the sum :
x+17 = 54
⇒ x = 54-17
⇒ x = 37
Required number = 37

Question 2.
A number decreased by 8 equals 26, find the number.
Solution:
Let required number = A
∴According to the sum :
x – 8 = 26
⇒A = 26 + 8
⇒A = 34
∴Required number = 34

Question 3.
One-fourth of a number added to two- seventh of it gives 135; find the number.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 68

Question 4.
Two-fifths of a number subtracted from three-fourths of it gives 56, find the number.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 69

Question 5.
A number is increased by 12 and the new number obtained is multiplied by 5. If the resulting number is 95, find the original number.
Solution:

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 70

Question 6.
A number is increased by 26 and the new number obtained is divided by 3. If the resulting number is 18; find the original number.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 71

Question 7.
The age of a man is 27 years more than the age of his son. If the sum of their ages is 47 years, find the age of the son and his father.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 72

Question 8.
The difference between the ages of Gopal and his father is 26 years. If the sum of their ages is 56 years, find the ages of Gopal and his father.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 73

Question 9.
When two consecutive natural numbers are added, the sum is 31; find the numbers.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 75

Question 10.
When three consecutive natural numbers are added, the sum is 66, find the numbers.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 76

Question 11.
A natural number decreased by 7 is 12. Find the number.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 77

Question 12.
One fourth of a number added to one- sixth of itself is 15. Find the number.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 78

Question 13.
A whole number is increased by 7 and the new number so obtained is multiplied by 5; the result is 45. Find the number.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 79.

Question 14.
The age of a man and the age of his daughter differ by 23 years and the sum of their ages is 41 years. Find the age of the man.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 80

Question 15.
The difference between the ages of a woman and her son is 19 years and the sum of their ages is 37 years; find the age of the son.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 82

Question 16.
Two natrual numbers differ by 6 and sum of them is 36. Find the larger number.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 83

Question 17.
The difference between two numbers is 15. Taking the smaller number as x; find:
(i) the expression for larger number.
(if) the larger number, if the sum of these numbers is 71.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 84

Question 18.
The difference between two numbers is 23. Taking the larger number as x, find:
(i) the expression for smaller number.
(ii) the smaller number, if the sum of these two numbers is 91.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 86

Question 19.
Find three consecutive integers such that their sum is 78.
Solution:
Sum of three consecutive numbers = 78
Let first number = x
Then second number = x + 1
and third number = x + 2
Then x + x+1+x + 2 = 78
⇒ 3x + 3 = 78
⇒ 3x = 78 – 3 = 75
⇒ x = \(\frac { 75 }{ 3 }\) =25
∴First number=25
Second number = 25 + 1 = 26
and third number = 26 + 1 = 27
Then the three required numbers are 25, 26,27

Question 20.
The sum of three consecutive numbers is 54. Taking the middle number as x, find:
(i) expression for the smallest number and the largest number.
(ii) the three numbers.
Solution:
Sum of three consecutive numbers = 54
Middle number = x
(i) The first number = x – 1
and third number = x + 1
(ii) ∴x + x-1+x+1 = 54
⇒ 3x = 54
⇒ x= \(\frac { 54 }{ 3 }\) = 18
∴First number =18-1 = 17
and third number =18 + 1 = 19
∴Three required numbers are 17, 18,19

Simple (Linear) Equations Revision Exercise – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Solve each of the following equations :
Question i.
2x + 3 = 7
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 87

Question ii.
2x – 3 = 7
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 88

Question iii.
2x ÷ 3 = 7
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 89

Question iv.
3x – 8 = 13
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 90

Question v.
3y + 8 = 13
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 91

Question vi.
3y ÷ 8 = 13
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 92

Question vii.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 93

Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 94

Question viii.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 95
Solution:

Question ix.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 98
Solution:

Question x.
5x – 2.4 = 4.9
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 100

Question xi.
5y + 4.9 = 2.4
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 101

Question xii.
48 z + 3.6 = 1.2
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 103

Question xiii.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 104
Solution:

Question xiv.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 106
Solution:

Question xv.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 109
Solution:

Question xvi.
-3x + 4 = 10
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 111

Question xvii.
5 = x – 3
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 112

Question xviii.
8y = 3- 3y
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 113

Question xix.
4x = 4.9 = 6.5
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 114

Question xx.
3z + 2 = -4
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 116

Question xxi.
7y -18 = 17
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 117

Question xxii.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 118
Solution:

Question xxiii.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 120
Solution:

Question xxiv.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 122
Solution:

Question xxv.
7x -2 = 4x +7
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 124

Question xxvi.
3y -(y -+2) =4
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 126

Question xxvii.
3z – 18 = z – (12 -4z)
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 127

Question xxiii.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 128
Solution:

Question xxix.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 131
Solution:

Question xxx.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 133
Solution:

Question xxxi.
5x – 2x +15 = 27
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 136

Question xxxii.
5y – 15 = 27 -2y
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 137

Question xxxiii.
7z + 15 = 3z – 13
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 138

Question xxxiv.
2 (x -3) – 3 (x-4) =12
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 139

Question xxxv.
(7y +8) +7= 8
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 141

Question xxxvi.
2(z-5) +3 (z+2) -(3-5z) =10
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 142

Question 2.
A natural number decreased by 7 is 12. Find the number.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 143

Question 3.
One-fourth of a number added to one-sixth of It is 15. Find the number.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 144

Question 4.
A whole number is increased by 7 and the number so obtained is multiplied by 5; the result is 45. Find the whole number.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 145

Question 5.
The age of a man and the age of his daughter differ by 23 years and the sum of their ages is 41 years. Find the age of the man.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 146

Question 6.
The difference between the ages of a woman and her son is 19 years and the sum of their ages is 37 years; find the age of the son.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 147

Question 7.
Two natural numbers differ by 6 and their sum is 36. Find the larger number.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 149

Question 8.
The difference between two numbers is 15. Taking the smaller number as x; find :
(i) the expression for the larger number.
(ii) the larger number, if the sum of these numbers is 71.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 150

Question 9.
The difference between two numbers is 23. Taking the larger number as x, find :
(i) the expression for smaller number.
(ii) the smaller number, if the sum of these two numbers is 91.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 151

Question 10.
Find the three consecutive integers whose sum is 78.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 153

Question 11.
The sum of three consecutive numbers is 54. Taking the middle number as x, find :
(i) the expressions for the smallest number and the largest number.
(ii) the three numbers.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations image - 154

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 23 Fundamental Concepts

August 11, 2022May 31, 2018 by Prasanna

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 23 Fundamental Concepts

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 23 Fundamental Concepts

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

Selina Class 6 Maths ICSE Solutions Physics Chemistry Biology Geography History & Civics

IMPORTANT POINTS

1. Fundamental Concepts : Geometry is the study of position,, shape, size and other properties of different figures. The geometrical terms such as : point, line, plane, etc., contain the basic ideas for the development of geometry.
(i) Point : A point is a mark of position. It has neither length nor width nor. thickness and occupies no space.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 23 Fundamental Concepts image - 1
(ii) Line : A line has only length. It has neither width nor thickness.
(iii) Ray : It is a line (i.e. a straight line) that starts from a given fixed point and moves in the same direction.

(iv) Line Segment: A line segmeftt is a part of a straight line. A line segment is a part of a line and also of a ray.

(v) Surface : A surface has length and width, but no thickness.
(vi) Plane : It is a flat surface. A plane has length and width, but no thickness.
(vii) Parallel Lines : Two straight lines are said to be parallel to each other if they lie in the same plane and do not meet when produced on either side.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 23 Fundamental Concepts image - 5
(viii)Intersecting Lines : If two lines lie in the same plane and are not parallel to each other, they are called intersecting lines.
(xi) Collinearity of Points : If three of more points lie on the same straight line, then the points are called collinear points.

(x) Concurrent Lines : If three or more straight lines pass through the same point, the lines are called concurrent lines.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 23 Fundamental Concepts image - 7Selina Concise Mathematics Class 6 ICSE Solutions Chapter 23 Fundamental Concepts image - 7

Fundamental Concepts Exercise 23A – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
State, true or false, if false, correct the statement.
(i) A dot has width but no length.
(ii) A ray has an infinite length only on one side of it.
(iii) A line segment PQ is written as \(\overleftrightarrow { PQ }\) .
(iv) \(\overleftrightarrow { PQ }\) represents a straight line.
(v) Three points are said to be collinear, if they lie in the same plane.
(vi) Three or more points all lying in the same line, are called collinear points.
Solution:
(i) False : Because a dot has no length, no breadth.
(ii) True.
(iii) False : A line segment PQ is written as PQ.
(iv) True.
( v) False: Three points are called collinear points if they are in the same straight line.
(vi) True.

Question 2.
Write how many lines can be drawn through :
(i) a given point ?
(ii) two given fixed points ?
(iii) three collinear points ?
(iv) three non-collinear points ?
Solution:
(i) Infinite (unlimited) line can be drawn through a given point.
(ii) only one line can be drawn through two given point.
(iii) only one line can be drawn through three collinear points.
(iv) None (no) line can be drawn through three non-collinear points.

Question 3.
The shaded region of the given figure shows a plane :
(a) Name :
(i) three collinear points.
(ii) three non-collinear points.
(iii) a pair of intersecting lines.
(b) State whether true or false :
(i) Line DE is contained in the given plane P.
(ii) Lines AB and DE intersect at point C.
(iii) Points D, B and C are collinear.
(iv) Points D, B and E are collinear.
Solution:
(a) (i) A, B and C are three collinear points.
(ii) A, D and C are non-collinear points.
(iii) AC and DE are intersecting lines.
(b) (i) True
(ii) True
(iii) False
(iv) False
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 23 Fundamental Concepts image - 8

Question 4.
Correct the statement, if it is wrong:
(i) A A ray can be extended infinitely on either side.
(ii) A ray has a definite length.
(iii) A line segment has a definite length.
(iv) A line has two end-points.
(v) A ray has only one end point.
Solution:
(i) A ray can be extended infinitely on one side of it only.
(ii) A ray has infinite length.
(iii) Yes, a line segment has a definite length.
(iv) A line-segment has two end-points.
(v) Yes, a ray has only one end-point.

Question 5.
State true-er false, if false give the correct statement :
(i) A line has a countable number of points in it.
(ii) Only one line can pass through a given point.
(iii) The intersection of two planes is a straight line
Solution:
(i) False, a line has length only.
(ii) False, any number of line can pass through a given point.
(iii) True.

Question 6.
State, whether the following pairs of lines or rays appear to be parallel or intersecting.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 23 Fundamental Concepts image - 9

Solution:
(i) intersecting
(ii) Parallel
(iii) Parallel
(iv) Intersecting

Question 7.
Give two examples, from your surroundings, for each of the following:
(i) points
(ii) line segments
(iii) plane surfaces
(iv) curved surfaces.
Solution:
(i) Tips of your pencil (Ball Pen) and Tip of paper pin.
(ii) Lines of Exercise Note-Books and edge of school desk.
(iii) Floor of the room and top of the table.
(iv) Surface of foot-ball and front glass of the car.

Question 8.
Under what condition will two straight
lines, in the same plane, have :
(i) no point in common.
(ii) only one point in common.
(iii) an infinite number of points in common.
(iv) If possible draw diagrams in support of your answer.
Solution:
(i) When lines are parallel to each others.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 23 Fundamental Concepts image - 11
(ii) When they intersect each other

Here, common point is E.
(iii) When line coincide with each other.

Question 9.
Mark two points A and B on a page of your exercise book. Mark a third point P, such that :
(i) P lies between A and B ; and the three points A, P and B are collinear.
(ii) P does not lie between A and B yet the three points are collinear.
(iii) the three points do not lie in a line.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 23 Fundamental Concepts image - 14

Question 10.
Mark two points P and Q on a piece of paper. How many lines can you draw:
(i) passing through both the points P and Q?
(ii) passing through the point P ?
(iii) passing through the point Q ?
Solution:
(i) From above diagram it is clear only one line can be drawn.

(ii) Infinite lines can pass through the point up
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 23 Fundamental Concepts image - 16
(iii) Infinite lines can pass through the point

Question 11.
The adjoining diagram shows a line AB. Draw diagram to represent:
(i) ray AB i.e. \(\xrightarrow { AB }\)

(ii) \(\xrightarrow { BA }\)
(iii) line segment AB.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 23 Fundamental Concepts image - 19

Question 12.
The adjoining diagram shows a ray AB. Draw diagrams to show :
(i) ray BA i.e. \(\xrightarrow { BA }\)
(ii) lineAB
(iii) line segment BA.

Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 23 Fundamental Concepts image - 21

Question 13.
The adjoining diagram shows a line segment AB. Draw diagrams to represent :
(i) ray AB i.e. \(\xrightarrow { AB }\)
(ii) line AB i.e. \(\overleftrightarrow { AB }\)
(iii) ray BA.

Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 23 Fundamental Concepts image - 23

Question 14.
Use a ruler and And whether following points are collinear or not:
(i) D, A and C
(ii) A, B and C
(iii) A, B and E
(iv) B, C and E
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 23 Fundamental Concepts image - 24
Solution:

Question 15.
From the adjoining figure, write:
(i) all pairs f parallel lines
(ii) all the lines which intersect EF.
(iii) lines whose point of intersection is G.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 23 Fundamental Concepts image - 26
Solution:
(i) The pairs of parallel lines are EF || GH, EF || IJ and GH || IJ
(ii) The lines which intersect EF are AB and CD
(iii) AB and GH are the lines whose point of intersection is G.

Fundamental Concepts Exercise 23B – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
State, which of the following is a plane closed figure :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 23 Fundamental Concepts image - 27
Solution:
(i) , (iii) ,(iv),(vi)and (viii) are plane close figures.

Question 2.
Fill in the blanks :
(i) ….. is a three sided plane closed figure.
(ii) A square is a plane closed figure which is not bounded by …..
(iii) A rectangle is a …. sided plane ……
(iv) A rectangle has opposite sides ….. and adjacent sides ….. to each other.
(v) The sides of a square are ……. to each other and each angle is
(vi) For a line segment, a line making angle of ……… with it, is called perpendicular to it.
(vii) For a line segment, a line ……. it and making angle of ……. with it, is called ……. bisector of the line segment.
(viii) How many perpendiculars can be drawn to a line segment of length 6 cm ?
(ix) How many perpendicular bisectors can be drawn to a line segment of length 6 cm ?
(x) A perpendicular to a line segment will be its perpendicular bisector if it passes through the ……. of the given line segment.
Solution:
(i) Triangle
(ii) any curved line.
(iii) four, closed figure.
(iv) parallel, perpendicular.
(v) Equal, 90°
(vi) 90°
(vii) bisecting, 90°, perpendicular.
(viii) Infinite.
(ix) Only one.
(x) mid-point

Question 3.
State, which of the lines/line- segments are perpendicular to the line PQ :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 23 Fundamental Concepts image - 28
Solution:
CD and MN are perpendicular to the line PQ. forming angle to 90°

Question 4.
Which of the following figures show two mutually perpendicular lines :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 23 Fundamental Concepts image - 29
Solution:
figures (i) and (iii) show two mutually perpendiculars lines forming angles of 90°.

Question 5.
For each figure given below name the line segment that is perpendicular bisector of the other :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 23 Fundamental Concepts image - 30
Solution:
(i) AB of CD
(ii) AB to MN
(iii) PQ to RS and RS to PQ
(iv) None
(v) None.

Question 6.
Name three objects from your surroundings that contain perpendicular edges.
Solution:
(i) My Text-Book of Mathematics r
(ii) My Note-Book
(iii) My Scale or My school’s Black Board

Question 7.
Using the given figure, answer the following :
(i) Name the pairs of parallel lines.
(ii) Name the pairs of mutually perpendicular lines.
(iii) Is the line p parallel to the line l ?
(iv) Is the line q perpendicular to the line m ?
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 23 Fundamental Concepts image - 31
Solution:
(i) l || m and p || q
(ii) p and l, p and m, q and l, q and m.
(iii) No, p is perpendicular to line l.
(iv) Yes.

Question 8.
Place a scale (ruler) on a sheet of paper and hold it firmly with one hand. Now draw two line segments AB and CD along the longer edges of the scale. State whether segment AB is parallel to or perpendicular to segment CD.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 23 Fundamental Concepts image - 32

Question 9.
Check your text book :
(i) How many pairs of its edges are parallel to each other ?
(ii) How many pairs of its edges are perpendicular to each other ?
Solution:
(i) 8
(ii) Three at each corner = 24

Question 10.
Give two examples from your surroundings for each of the following :
(i) intersecting lines
(ii) parallel lines
(iii) perpendicular lines.
Solution:
(i) Edges of my Text-Book and Note Book through any comer.
(ii) opp. edges of my Text-Book and Note Book.
(iii) Adjacent edges of my Text-Book and Black Board in my class-room.

Question 11.
State true or false; if false, give the correct statement :
(i) The maximum number of lines through three collinear points is three.
(ii) The maximum number of lines through three non-collinear points is three.
(iii) Two parallel lines always lie in the same plane.
(iv) Concurrent lines always meet at the same point.
(v) A surface can be plane or curved.
(vi) There are an infinite number of points in a line segment of length 10 cm.
(vii) There are an infinite number of points in a line.
(viii) A plane has an infinite number of lines.
(ix) A plane has an infinite number of points.
Solution:
(i) False: Because only one line passes through three collinear points.
(ii) False : No line passes through all the three non-collinear points.
(iii) True.
(iv) True.
(v) True.
(vi) True.
(vii) True.
(viii) True.
(ix) True.

Fundamental Concepts Revision Exercise – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Mark a point O on a piece of paper. Using a sharp pencil and a ruler, draw a line through the point O. Then draw one more line through the point O. How many lines can you draw through O?
Write a statement regarding your observations.
Solution:
We take a point O on a piece of paper. Using a sharp pencil, we draw a line OA through O

another line OB is drawn as shown in the figure.
We can draw an infinite number of lines from O as we know that an infinite number of lines can be drawn through a point in aplane,

Question 2.
Mark two points A and B on a plain sheet “ of paper. Using a sharp pencil and a ruler, draw a line through points A and B. Try to draw one more line through A and B. What do you observe ?
Write a statement regarding your observations.
Solution:
Two points A and B are marked On the plain sheet of paper. Using pencil, draw a line through there two points.

We cannot draw any other line joining them as only one line can be drawn passing through two fixed (given) points in a plane.

Question 3.
Draw two straight lines on a sheet of paper so that the lines drawn :
(i) intersect each other.
(ii) do not intersect.
(iii) appear to intersect when extended.
Solution:
Two straight lines are drawn on the sheet of a paper
(i) Intersecting each other:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 23 Fundamental Concepts image - 35
(ii) Not intersecting each other:

(iii) Intersect each other when produced them:

Question 4.
Draw a figure to show that :
(i) points A, B and C arc collinear.
(ii) lines AB, CD and EF are concurrent.
Solution:
(i) We know that three or more points are collinear if they lie on the same straight line. Therefore the required figure (line) is given below on which three points A, B and C lie

(ii) If three or more lines pass through a common point, these are called concurrent lines. Below is given the figure, in which three lines AB, CD and EF are passing (or intersecting each other) at O.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 23 Fundamental Concepts image - 39

Question 5.
In your note-book, draw two rays with the same initial point O and going in opposite directions. Write the special name for the final figure obtained.
Solution:
Two rays OA and OB are drawn through O in opposite direction as given below.

There two rays form a straight line \(\overleftrightarrow { AB }\)

Question 6.
In each figure, given below, write the number of line segments used :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 23 Fundamental Concepts image - 41
Solution:
(i) In the given figure, there as Ten (10) line segments which are AB, BC, CD, DE, EF, FG, GH, HI, IJ and JA.

(ii) In the given figure, there are 12 line segments which are AB, BC, CD, DE, EF, FG, GH, HI, IJ, JK, KL and LA.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 23 Fundamental Concepts image - 43
(iii) In the given figure, there are 13 line segments which are AB, BC, CD, DE, EF, FG, GH, HI, IJ, JK, KL, LM and MA.

(iv) In the given figure, there are 7 line segments which are AB, BC, CD, DE, DA, DB, CE.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 23 Fundamental Concepts image - 45

Question 7.
In a circle, point P is its centre and PA = PB = radius of the circle. Where do points A and B lie ?
Solution:
A circle is given with centre P and PA = PB = radius of the circle.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 23 Fundamental Concepts image - 46
∵ The radii of a circle are equal there they lie on the circumference of the circle In other words, path of a point which is equidistant from a fixed point is the circumference of a circle. The fixed point is called the centre and equidistance is the radius.

Question 8.
Draw a four-sided closed figure, in which:
(i) all the sides are equal and each angle is 90°.
(ii) opposite sides are equal and each angle is 90°.
Write the special name of each figure drawn.
Solution:
(i) In this four sided closed figure all the sides are equal and each angle is 90°. This figure is called a square. In ABCD, AB = BC = CD = DA and ∠A = ∠B =∠C = ∠D = 90°
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 23 Fundamental Concepts image - 47
(ii) In this four sided closed figure opposite sides are equal and each angle is 90°. This figure is called a rectangle In rectangle ABCD, AB = CD and BC = AD
∠A = ∠B = ∠C = ∠D = 90°.

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles

August 11, 2022May 31, 2018 by Prasanna

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles (With their Types)

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles (With their Types)

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IMPORTANT POINTS
1. Ray. A ray is a half-line: It has one end point and other end is open. It can not be measured like a line. Here, OA is a ray.

From appoint, infinite numbers of rays can be drawn.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 2
2. Angle: When two rays meet at a point, then an angle is formed. Angle is measured in degrees with the help of an instrument known as a protractor.
The point where the two rays meet is called an initial point or vertex of the angle and the two rays which form the angle, are called the sides of the angle e.g., two rays OA and OB meet at O.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 3
∴ Angle AOB is formed. Vertex is always kept between the ends points.
The anlgle can be denotes as ∠AOB, or ∠BOA, here sign ‘∠’ denotes the angle.
The angle is also denoted with letters A, B, C etc. or numbers 1, 2, 3 etc. also.

3. Parts of an angl: Angle has three parts : Interior, exterior and the angle itself.
4. Comparison of Angles: Two angles can be compared with respect to their magnitude. Any angle of greater measure is greater.

5. Kinds of Angles :
(i) Zero angle: When two rotating rays (sides of angles) coincide each other, then ∠ero angle is formed.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 4
(ii) Right angle: An angle of 90° is called a right angle.

(iii) Straight Angle: An angle of 180° is called a straight angle.

(iv) Complete Angle: When a ray completes a revolution on rotating it, then a complete angle is formed.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 7
(v) Acute angle : An angle less than 90° is called an acute angle.

(vi) Obtuse angle : An angle greater than 90° and less than 180° is called an obtuse angle.

(vii) Reflex Angle: An angle greater than the 180° and less than 360°
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 10
Note:
1 ° = 60 minutes (60′)
1′ = 60 seconds (60″)

6. Pairs of angles :
(i) Adjacent Angles: Two angles with same vertex and one common arm and the other arms lying in opposite sides of it are called adjacent angles, ∠AOB and ∠BOC are adjacent angles.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 11
(ii) Linear Pair: A linear pair is a pair of adjacent angles whose sum is equal to 180°. ∠AOB and ∠BOC are a linear pair as ∠AOB + ∠BOC = 180°.

(iii) Complementary Angle: Two angles whose sum is 90° are called complementary angles. ∠ABC and ∠PQR are complementary angles as ∠ABC + ∠PQR = 30° + 60° = 90°.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 13
(iv) Supplementary Angles: Two angles whose sum is 180°, are called supplementary angles. ∠ABC and ∠PQR are supplementary angles, because
∠ABC + ∠PQR = 130° + 50° = 180°.

(v) Vertically Opposite Angles: When two lines intersect each other, then the pairs of opposite angles so formed are called vertically opposite angles.
∠1 and ∠2 are vertically opposite angles. Similarly, ∠3 and ∠4 are vertically opposite angles.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 15

Angles Exercise 24A – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
For each angle given below, write the name of the vertex, the names of the arms and the name of the angle.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 16
Solution:
(i) In figure (i) O is the vertex, OA, OB are its arms and name of the angle is ∠AOB or∠BOA or simply ∠O.
(ii) In figure (ii) Q is the vertex, QP and QR its arms and the name of the angle is ∠PQR or ∠RQP or simply ∠Q.
(iii) In figure (iii), M is the vertex, MN and ML and its anus, and name of the angle is ∠LMN or ∠NML or simply ∠M.

P .Q . Name the angles marked by letters a, b, c, x and y.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 17

Solution:
a = AOE, b = ∠AOB, c = ∠BOC d = ∠COD e= ∠DOE

Question 2.
Name the points :
(i) in the interior of the angle PQR,
(ii) in the exterior of the angle PQR.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 18
Solution:
(i) a, b and x
(ii) d, m, n, s, and t.

Question 3.
In the given figure, figure out the number of angles formed within the arms OA and OE.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 19
Solution:
∠AOE, ∠AOD, ∠AOC, ∠AOB, ∠BOC ∠BOD, ∠BOE, ∠COD, ∠COE and ∠DOE.

Question 4.
Add :
(i) 29° 16′ 23″ and 8° 27′ 12″
(ii) 9° 45’56” and 73° 8′ 15″
(m) 56° 38′ and 27° 42’30”
(iv) 47° and 61° 17’4″
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 20

Question 5.
In the figure, given below name :
(i) three pairs of adjacent angles.
(ii) two acute angles,
(iii) two obtuse angles
(iv) two reflex angles.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 21
Solution:
(i) Three pairs of adjacent angles are
∠AOB and ∠BOC;
∠BOC and ∠COD;
∠COD and ∠DOA.
(ii) Two acute angles are
∠AOB and ∠AOD.
(iii) Two obtuse angles are
∠BOC and ∠COD.
(iv) Two reflex angles are
∠AOB and ∠COD.

Question 6.
In the given figure ; PQR is a straight line. If :
(i) ∠SQR = 75° ; find ∠PQS.
(ii) ∠PQS = 110°; find ∠RQS
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 22
Solution:
(i) From figure,
∠PQS + ∠SQR = 180° [Linear pair of angles]
⇒∠PQS + 75° = 180°
⇒ ∠PQS = 180°-75°
⇒ ∠PQS = 105°
(ii) From figure again,
∠PQS + ∠RQS = 180°
⇒ 110° + ∠RQS = 180°
∠RQS = 180°- 110°
∠RQS = 70°

Question 7.
In the given figure ; AOC-is a straight line. If angle AOB = 50°, angle AOE = 90° and angle COD = 25° ; find the measure of :
(i) angle BOC
(ii) angle EOD
(iii) obtuse angle BOD
(iv) reflex angle BOD
(v) reflex angle COE.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 23
Solution:
(i) ∠AOB + ∠BOC = 180° (Linear pairs of angle)
⇒ 50° +∠BOC = 180°
⇒ ∠BOC = 180° – 50° = 130°
⇒ ∠BOC = 130°
(ii) ∠EOD + ∠COD = 90° (∵AOE = 90°)
⇒ ∠EOD + 25° = 90°
⇒ ∠EOD + 25° = 90°
⇒ ∠EOD = 90° – 25°
⇒ ∠EOD = 65°
(iii) ∠BOD = ∠BOC + COD
= 130° + 25° = 155°
(iv) Reflex ∠BOD = 360° – ∠BOD
= 360°- 155° = 205°
(v) Reflex ∠COE = 360° – ∠COE
= 360° (∠COD + ∠EOD)
= 360° – (25° + 65°)
= 360° – 90° = 270°

Question 8.
In the given figure if :
(i) a = 130° ; find b.
(ii) b = 200 ; find a.
(iii) a = 5/3 right angle, find b
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 24
Solution:
(i) From figure, a + b = 360°
⇒ 130° + 6 = 360
⇒ 6 = 360°-130°
⇒ b = 230°
(ii) From figure,
a + b = 360°
⇒ a + 200° = 360°
⇒ a = 360° – 200°
⇒ a = 160°
(iii) Here, a= \(\frac { 5 }{ 3 }\) right angle
= \(\frac { 5 }{ 3 }\) x90° = 150°
a = 150°
Here, a + b = 360°
⇒ 150° + b = 360° (∵a = 150°)
⇒ b = 360° -150°
b = 210°

Question 9.
In the given diagram, ABC is a straight line.
(i) If x = 53°, find y.
(ii) If y =1\(\frac { 1 }{ 2 }\) right angles ; find x.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 25
Solution:
(i) From the figure,
∠ABD + ∠DBC = 180° (Linear pair of angles)
⇒ x+y=180°
⇒ 53°+y = 180° (∵ x = 53”)
⇒ y = 180° – 53°
⇒ y = 127°
(ii) From figure again,
x+y= 180°
1 + \(\frac { 3 }{ 2 }\) x 90 = 180°
⇒ x+1\(\frac { 1 }{ 2 }\) right angles = 180°
⇒ x+\(\frac { 3 }{ 2 }\) x90=180°
⇒ x + 135°= 180°
⇒ x= 180° – 135°
⇒ x = 45°

Question 10.
In the given figure, AOB is a straight line. Find the value ofx and also answer each of the following :
(i) ∠AOP = ……..
(ii) ∠BOP = ……..
(iii) which angle is obtuse ?
(iv) which angle is acute ?
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 26
Solution:
∠AOP = x + 30°
∠BOP = x – 30°
But ∠AOP + ∠BOP = 180° (∵ ∠AOB is a straight angle)
⇒ x + 30°+x-30° = 180°
⇒ 2x = 180°
⇒ x = 90°
(i) ∠AOP = x + 30° = 90° + 30° = 120°
(ii) ∠BOP = x- 30° = 90° – 30° = 60°
(iii) ∠AOP is an obtuse angle
(iv) ∠BOP is an acute angle

Question 11.
In the given figure, PQR is a straight line. Find x. Then complete the following:
(i) ∠AQB = ……..
(ii) ∠BQP = ……..
(iii) ∠AQR = …….
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 27
Solution:
PQR is a straight line
∠AQP=x + 20°
∠AQB = 2x + 10°
∠BQR = x – 10°
But ∠AQP + ∠AQB + ∠BQR = 180°
⇒ x + 20° + 2x + 10° + x-10°= 180°
⇒ 4x + 20°= 180°
⇒4x= 180°-20°= 160°
⇒ x = \(\frac { 160 }{ 4 }\)° = 40°
(i) ∠AQB = 2x + 10° = 2 x 40° + 10° = 80° + 10° = 90°
∠AQP = x + 2(T = 40° + 20° = 60°
∠BQR = x – 10° = 40° – 10° = 30°
(ii) ∠BQP = ∠AQP + ∠AQB = 60° + 90° = 150°
(iii) ∠AQR = ∠AQB + ∠BQR = 90° + 30° = 120°

Question 12.
In the given figure, lines AB and CD intersect at point O.
(i) Find the value of ∠a.
(ii) Name all the pairs of vertically opposite angles.
(iii) Name all the pairs of adjacent angles.
(iv) Name all the reflex angles formed and write the measure of each.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 28
Solution:
Two lines AB and CD intersect each other at O
∠AOC = 68°
(i) ∵ AOB is a line
∠AOC + ∠BOC = 180°
⇒ 68° + a = 180°
⇒ a= 180°-68° = 112°
(ii) ∠AOC and ∠BOD and ∠BOC and ∠AOD are the two pairs of vertically opposite angles .
(iii) ∠AOC and ∠BOC; ∠BOC and ∠BOD; ∠BOD and ∠DOA;
∠DOA and AOC are the pairs of adjacent angles
(iv) ∠BOC and ∠DOA are reflex angles and also ∠AOC and ∠BOD are also reflex angles
Ref. ∠BOC = 180° + 68° = 248°
Ref. ∠DOA = 180° + 68° = 248°
Ref. ∠AOC = 180° + 112° = 292°
and ref. ∠BOD =180° + 112° = 292°

Question 13.
In the given figure :
(i) If ∠AOB = 45°, ∠BOC = 30° and ∠AOD= 110°;
find : angles COD and BOD.
(ii) If ∠BOC = ∠DOC = 34° and ∠AOD = 120° ;
find : angle AOB and angle AOC.
(iii) If ∠AOB = ∠BOC = ∠COD = 38°
find : reflex angle AOC and reflex angle AOD.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 29
Solution:
(i) ∠COD = ∠AOD – ∠AOC
= ∠AOD – (∠AOB + ∠BOC)
= 110°-(45°+ 30°)
= 110°-75° = 35°
∠BOD = ∠AOD -∠AOB
= 110° -45°
= 65°
(ii) ∠AOB = ∠AOD-∠BOD
= ∠AOD – (∠BOC + ∠COD)
= 120° – (34° + 34°)
= 120°-68°
= 52°
∠AOC = ∠AOB + ∠BOC
= 52° + 34°
= 86°
(iii) Reflex ∠AOC = 360°-∠AOC
= 360° – (∠AOB + ∠BOC)
= 360° – (38° + 38°)
= 360° – 76° = 284°
Reflex ∠AOD = 360° – ∠AOD
= 360° (∠AOB + ∠BOC + ∠COD)
= 360° – (38° + 38° + 38°)
= 360°- 114°
= 246°

Angles Exercise 24B – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Write the complement angle of :
(i) 45°
(ii) x°
(iii) (x – 10)°
(iv) 20° + y°
Solution:
(i) Complement angle of 45°
= 90° – 45° = 45°
(ii) Complement angle of x°
= 90° – x° = (90 – x)°
(iii) Complement angle of (x – 10)° = 90° (x – 10°)
= 90°-x + 10° = 100°-x
(iv) Complement angle of 20° + y°
= 90°-(20°+y°)
= 90° – 20° -y° = 70° -y°

Question 2.
Write the supplement angle of :
(i) 49°
(ii) 111°
(iii) (x – 30)°
(iv) 20° + y°
Solution:
(i) Supplement angle of 49°
= 180°-49° = 131°
(ii) Supplement angle of 111°
= 180°- 1110 = 69°
(iii) Supplment of (x – 30)° = 180° – ( x° – 30°)
= 180o – x° + 30° – 210° – x°
(iv) Supplement of ∠20° + y° = 180° – (20° +y°)
= 180° -20°-y°
= 160° -y°

Question 3.
Write the complement angle of :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 30

Solution:

Question 4.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 33
Solution:

Question 5.
Find the angle :
(i) that is equal to its complement ?
(ii) that is equal to its supplement ?
Solution:
(i) 45° is equal to its complement.
(ii) 90° is equal to its supplement.

Question 6.
Two complementary angles are in the ratio 7 : 8. Find the angles.
Solution:
Let two complementary angles are lx and 8x
∴ 7x + 8x = 90°
⇒ 15x = 90°
⇒ x = \(\frac { 90 }{ 15 }\)°
⇒ x = 6°
∴Two complementary angles are
7x = 7 x 6° = 42°
8x = 8 x 6° = 48°

Question 7.
Two supplementary angles are in the ratio 7 : 11. Find the angles.
Solution:
Let two supplementary angles are 7x and 11x
∴ 7x+ 11x= 180°
⇒ 18x = 180°
⇒ x = \(\frac { 180 }{ 18 }\)
⇒ x = 10°
Two supplementary angles are
7x = 7 x 10° = 70°
11x= 11 x 10°= 110°

Question 8.
The measures of two complementary angles are (2x – 7)° and (x + 4)°. Find x.
Solution:
We know that, sum of two complementary angles = 90°
∴(2x – 7) + (x + 4) = 90°
2x-7 + x + 4 = 90°
⇒ 2x + x – 7 + 4 = 90°
⇒ 3x – 3 = 90°
⇒ 3x = 90 + 3
⇒ 3x = 93
⇒ x = \(\frac { 93 }{ 3 }\)
x = 31

Question 9.
The measures of two supplementary angles are (3x + 15)° and (2x + 5)°. Find x.
Solution:
We know that, sum of two supplementary angles = 180°
∴ (3x + 15)° + (2x + 5)° = 180° ‘
3x + 15 + 2x + 5 = 180°
⇒ 3x + 2x+15 + 5 = 180°
⇒ 5x°+ 20° = 180°
⇒ 5x = 180° – 20°
⇒ 5x= 160°
⇒ x = \(\frac { 160 }{ 5 }\)
⇒ x = 32°

Question 10.
For an angle x°, find :
(i) the complementary angle
(ii) the supplementary angle
(iii) the value of x° if its supplementary angle is three times its complementary angle.
Solution:
For an angle x,
(i) Complementary angle of x° = (90° – x)
(ii) Supplementary angle of x° = (180° – x)
(iii) ∵‘Supplementary angle = 3 (complementary anlge)
180°- x = 3 (90°-x)
⇒ 180°-x = 270°- 3x
⇒-x + 3x = 270°-180°
⇒ 2x = 90°
⇒x= \(\frac { 90 }{ 2 }\) = 45°
∴ x = 45°

Angles Revision Exercise – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Explain what do you understand by :
(i) Adjacent angles ?
(ii) Complementary angles ?
(iii) Supplementary angles ?
Solution:
(i) Adjacent Angles: Two angles are called adjacent angles if (a) they have a common vertex (b) they have one common arm and (iii) the other two arms of the angles are on the opposite sides of the common arm.
(ii) Complementary Angles : Two angles whose sum is 90° are called complementary angles to each other.
(iii) Supplementary Angles : Two angles whose sum.is 180° are called supplementary angles to each other.

Question 2.
Find the value of ‘x’ for each of the following figures :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 36

Solution:

Question 3.
Find the number of degrees in an angle that is (i) \(\frac { 3 }{ 5 }\) of a right angle (ii) 0.2 times of a straight line angle.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 40

Question 4.
In the given figure; AB, CD and EF are straight lines. Name the pair of angles forming :
(i) straight line angles.
(ii) vertically opposite angles.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 41
Solution:
In the given figure, AB, CD and EF are straight lines on intersecting, angles are formed a, b, c, d, l, m, n and p.
(i) In the figure pairs of straight line angles are ∠a, ∠b; ∠b, ∠c; ∠c, ∠d; ∠d, ∠a
∠l, ∠m; ∠m, ∠n;∠n, ∠p and ∠p, ∠l
(ii) Pairs of vertically angles are ∠a, ∠c; ∠b, ∠d; ∠l, ∠n; ∠m, ∠p

Question 5.
Find the complement of :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 42
Solution:

Question 6.
Find the supplement of :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 44.
Solution:

Question 7.
Two complementary angles are in the ratio 8 : 7. Find the angles.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 46

Question 8.
Two supplementary angles are in the ratio 7 : 5. Find the angles.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 47

Question 9.
Two supplementary angles are (5x – 82°) and (4x + 73°). Find the value of x.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 48

Question 10.
Find the angle formed by the arms of a clock at:
(i) 3 O’clock
(ii) 6 O’clock
(iii) 9 O’clock
(iv) 12 O’clock
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 49

Question 11.
For an angle y°, find :
(i) its supplementary angle.
(ii) its complementary angle.
(iii) the value of y° if its supplement is four times its complement.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 51

Question 12.
Use the adjoining figure to find :
(i) ∠BOD
(ii) ∠AOC
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 52

Question 13.
Two adjacent angles forming a linear pair are in the ratio 7:5, find the angles.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 53

Question 14.
Find the angle that is three times its complementary angle.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 54

Question 15.
An angle is one-thirds of a straight line angle ; find :
(i) the angle
(ii) the complement and the supplement of the angle obtained above.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 24 Angles image - 55

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines

August 11, 2022May 30, 2018 by Prasanna

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines (Including Parallel Lines)

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines (Including Parallel Lines)

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

Selina Class 6 Maths ICSE Solutions Physics Chemistry Biology Geography History & Civics

IMPORTANT POINTS
1. Property : When two straight lines intersect:
(i) sum of each pair of adjacent angles is always 180°.
(ii) vertically opposite angles are always equal. .
2. Property : If the sum of two adjacent angles is 180°, their exterior arms are always in the same straight line.
Conversely, if the exterior arms of two adjacent angles are in the same straight line ; the sum of angles is always 180°
3. Parallel Lines : Two straight lines are said to be parallel, if they do not meet anywhere, no matter how much they are produced in either direction.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 1
4. Concepts of Transversal Lines : When a line cuts two or more lines (parallel or non-parallel); it is called a transversal line or simply, a transversal. In each of the following figures : PQ is a transversal line.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 2
5. Angles formed by two lines and their transversal line : When a transversal cuts two parallel or nonparallel lines; eight (8) angles are formed which are marked 1 to 8 in the adjoining diagram.
These angles can further he distinguished, as given below:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 3
(i) Exterior Angles : Angles marked 1, 2, 7 and 8 are exterior angles.
(ii) Interior Angles : Angles marked 3, 4, 5 and 6 are interior angles.
(iii) Exterior Alternates Angles : Two pairs of exterior alternate angles are marked as : 2 and 8 ; and, 1 and 7.
(iv) Interior Alternate Angles : Two pairs of interior alternate are marked as : 3 and 5 ; and 4 and 6. In general, interior alternate angles are simply called as alternate angles only.
(v) Corresponding Angles : Four pairs of corresponding angles are marked as : 1 and 5 ; 2 and 6 ; 3 and 7 ; and 4 and 8.
(vi) Co-interior or Conjoined or Allied Angles : Two pairs of co-interior or allied angles are marked as : 3 and 6 ; and 4 and 5.
(vii) Exterior Allied Angles : Two pairs of exterior allied angles are marked as : 2 and 7 ; and 1 and 8.

Properties of Angles and Lines Exercise 25A – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Two straight lines AB and CD intersect each other at a point O and angle AOC = 50° ; find :
(i) angle BOD
(ii) ∠AOD
(iii) ∠BOC
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 4
Solution:
(i)∠BOD = ∠AOC
(Vertically opposite angles are equal)
∴ ∠BOD =50°
(ii) ∠AOD
∠AOD + ∠BOD = 180°
∠AOD + 50° = 180° [From (i)]
∠AOD = 180°-50°
∠AOD = 130°
(iii) ∠BOC = ∠AOD
(Vertically opposite angles are equal)
∴ ∠BOC =130°

Question 2.
The adjoining figure, shows two straight lines AB and CD intersecting at point P. If ∠BPC = 4x – 5° and ∠APD = 3x + 15° ; find :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 5
(i) the value of x.
(ii) ∠APD
(iii) ∠BPD
(iv) ∠BPC
Solution:

Question 3.
The given diagram, shows two adjacent angles AOB and AOC, whose exterior sides are along the same straight line. Find the value of x.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 6
Solution:
Since, the exterior arms of the adjacent angles are in a straight line ; the adjacent angles are supplementary
∴ ∠AOB + ∠AOC = 180°
⇒ 68° + 3x – 20° = 180°
⇒ 3x = 180° + 20° – 68°
⇒ 3x = 200° – 68° ⇒ 3x =132°
x = \(\frac { 132 }{ 3 }\)° = 44°

Question 4.
Each figure given below shows a pair of adjacent angles AOB and BOC. Find whether or not the exterior arms OA and OC are in the same straight line.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 7

Solution:
(i) ∠AOB + ∠COB = 180°
Since, the sum of adjacent angles AOB and COB = 180°
(90° -x) + (90°+ x) = 180°
⇒ 90°-x + 90° + x = 180°
⇒ 180° =180°
The exterior arms. OA and OC are in the same straight line.
(ii) ∠AOB + ∠BOC = 97° + 83° = 180°
⇒ The sum of adjacent angles AOB and BOC is 180°.
∴ The exterior arms OA and OC are in the same straight line.
(iii)∠COB + ∠AOB = 88° + 112° = 200° ; which is not 180°.
⇒ The exterior amis OA and OC are not in the same straight line.

Question 5.
A line segment AP stands at point P of a straight line BC such that ∠APB = 5x – 40° and ∠APC = .x+ 10°; find the value of x and angle APB.
Solution:
AP stands on BC at P and
∠APB = 5x – 40°, ∠APC = x + 10°
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 9
(i) ∵APE is a straight line
∠APB + ∠APC = 180°
⇒ 5x – 40° + x + 10° = 180°
⇒ 6x-30°= 180°
⇒6x= 180° + 30° = 210°
x = \(\frac { 210 }{ 6 }\)° = 35°
(ii) and ∠APB = 5x – 40° = 5 x 35° – 40°
= 175 ° – 140° = 135°

Properties of Angles and Lines Exercise 25B – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Identify the pair of angles in each of the figure given below :
adjacent angles, vertically opposite angles, interior alternate angles, corresponding angles or exterior alternate angles.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 10

Solution:
(a) (i) Adjacent angles
(ii) Alternate exterior angles
(iii) Interior alternate angles
(iv) Corresponding angles
(v) Allied angles
(b) (i) Alternate interior angles
(ii) Corresponding angles
(iii) Alternate exterior angles
(iv) Corresponding angles
(v) Allied angles.
(c) (i) Corresponding
(ii) Alternate exterior
(iii) Alternate interior
(iv) Alternate interior
(v) Alternate exterior
(vi) Vertically opposite

Question 2.
Each figure given below shows a pair of parallel lines cut by a transversal For each case, find a and b, giving reasons.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 12

Solution:

Question 3.
If ∠1 = 120°, find the measures of : ∠2, ∠3, ∠4, ∠5, ∠6, ∠7 and ∠8. Give reasons.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 17
Solution:

Question 4.
In the figure given below, find the measure of the angles denoted by x,y, z,p,q and r.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 19
Solution:

Question 5.
Using the given figure, fill in the blanks.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 21
Solution:

Question 6.
In the given figure, find the anlges shown by x,y, z and w. Give reasons.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 23
Solution:

Question 7.
Find a, b, c and d in the figure given below :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 25
Solution:

Question 8.
Find x, y and z in the figure given below :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 27
Solution:

Properties of Angles and Lines Exercise 25C – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
In your note-book copy the following angles using ruler and a pair compass only.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 29
Solution:
(i) Steps of Construction :
1. At point Q, draw line QR = OB.

2. With O as centre, draw an arc of any suitable radius, to cut the arms of the angle at C and D.
3. With Q as centre, draw the arc of the same size as drawn for C and D. Let this arc cuts line QR at point T.
4. In your compasses, take the distance equal to distance between C and D; and then with T as centre, draw an arc which cuts the earlier arc at S.
5. Join QS and produce upto a suitable point P. ∠PQR so obtained, is the angle equal to the given ∠AOB.
(ii) Steps of Construction :
1. A t point E, draw line EF.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 31
2. With E as centre, draw an arc of any suitable radius, to cut the amis of the angle at C and D.
3. With Q as centre, draw the arc of the same size as drawn for C and D. Let this arc cuts line QR at point T.
4. In your compasses, take the distance equal to distance between C and D ; and then with T as centre, draw an arc which cuts the earlier arc at S.
5. Join QS and produce upto a suitable point R ∠PQR, so obtained, is the angle equal to the given ∠DEE
(iii) Steps of Construction :
1. At point A draw line AB = QP
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 32
2. With Q as centre, draw an arc of any suitable radius, to cut the arms of the angle A + C and D.
3. With A as centre, draw the arc of the same size as drawn for C and D. Let this arc cuts line AB at D.
4. In your compasses, take the distance equal to distance between 7 and 5 ; and then with D as centre, draw an arc which cuts the earlier arc at E.
5. Join AE and produced upto a suitable point C. ∠BAC, so obtained is the angle equal to the given ∠PQR.

Question 2.
Construct the following angles, using ruler and a pair of compass only
(i) 60°
(ii) 90°
(iii) 45°
(iv) 30°
(v) 120°
(vi) 135°
(vii) 15°
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 33

Question 3.
Draw line AB = 6cm. Construct angle ABC = 60°. Then draw the bisector of angle ABC.
Solution:
Steps of Construction :
1. Draw a line segment AB = 6 cm.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 39
2. With the help of compass construct ∠CBA = 60°.
3. Bisect ∠CBA, with the help of compass, take any radius which meet line AB and BC at point E and F.
4. Now, with the help of compass take
radius more than \(\frac { 1 }{ 2 }\) of EF and draw two arcs from point E and F, which intersect both arcs at G, proceed BG toward D ∠DBA is bisector of ∠CBA.

Question 4.
Draw a line segment PQ = 8cm. Construct the perpendicular bisector of the line segment PQ. Let the perpendicular bisector drawn meet PQ at point R. Measure the lengths of PR and QR. Is PR = QR ?
Solution:
Steps of Construction :
1. With P and Q as centres, draw arcs on both sides of PQ with equal radii. The radius should be more than half the length of PQ.
2. Let these arcs cut each other at points R and RS
3. Join RS which cuts PQ at D.
Then RS = PQ Also ∠POR = 90°.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 40
Hence, the line segment RS is the perpendicular bisector of PQ as it bisects PQ at P and is also perpendicular to PQ. On measuring the lengths of PR = 4cm, QR = 4 cm
Since PR = QR, both are 4cm each
∴PR = QR.

Question 5.
Draw a line segment AB = 7cm. Mark a point Pon AB such that AP=3 cm. Draw perpendicular on to AB at point P.
Solution:
1. Draw a line segment AB = 7 cm.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 41
2. Out point from AB – AP =3cm
3. From point P, cut arc on out side of AB, E and F.
4. From pont E & F cut arcs on both side intersection each other at C & D.
5. Join point P, CD.
6. Which is the required perpendicular.

Question 6.
Draw a line segment AB = 6.5 cm. Locate a point P that is 5 cm from A and 4.6 cm from B. Through the point P, draw a perpendicular on to the line segment AB.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 42
Steps of Construction :
(i) Draw a line segment AB =6.5cm
(ii) With centre A and radius 5 cm, draw an arc and with centre B and radius 4.6 cm, draw another arc which intersects the first arc at P.
Then P is the required point.
(iii) With centre A and a suitable radius, draw an arc which intersect AB at E and F.
(iv) With centres E and F and radius greater than half of EF, draw the arcs which intersect each other at Q.
(v) Join PQ which intersect AB at D.
Then PD is perpendicular to AB.

Properties of Angles and Lines Exercise 25D – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Draw a line segment OA = 5 cm. Use set-square to construct angle AOB = 60°, such that OB = 3 cm. Join A and B ; then measure the length ofAB.
Solution:
Measuring the length of AB = 4.4cm. (approximately)
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 43

Question 2.
Draw a line segment OP = 8cm. Use set-square to construct ∠POQ = 90°; such that OQ = 6 cm. Join P and Q; then measure the length of PQ.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 44

Question 3.
Draw ∠ABC = 120°. Bisect the angle using ruler and compasses. Measure each angle so obtained and check whether or not the new angles obtained on bisecting ∠ABC are equal.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 45

Question 4.
Draw ∠PQR = 75° by using set- squares. On PQ mark a point M such that MQ = 3 cm. On QR mark a point N such that QN = 4 cm. Join M and N. Measure the length of MN.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 46

Properties of Angles and Lines Revision Exercise – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
In the following figures, AB is parallel to CD; find the values of angles x, y and z :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 47
Solution:

Question 2.
In each of the following figures, BA is parallel to CD. Find the angles a, b and c:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 51
Solution:

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 53

Question 3.
In each of the following figures, PQ is parallel to RS. Find the angles a, b and c:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 54
Solution:

Question 4.
Two straight lines are cut by a transversal. Are the corresponding angles always equal?
Solution:
If a transversal cuts two straight lines, their the corresponding angles are not equal unless the lines are not parallel. One in case of parallel lines, the corresponding angles are equal.

Question 5.
Two straight lines are cut by a transversal so that the co-interior angles are supplementary. Are the straight lines parallel ?
Solution:
A transversal intersects two straight lines and co-interior angles are supplementary
∴ By deflations, the lines will be parallel.

Question 6.
Two straight lines are cut by a transversal so that the co-interior angles are equal. What must be the measure of each interior angle to make the straight lines parallel to each other ?
Solution:
A transveral intersects two straight lines and co-interior angles are equal to each other,
∵ The two straight lines are parallel Their sum of co-interior angles = 180°
But both angles are equal
∴ Each angle will be \(\frac { 180 }{ 2 }\)° = 90°

Question 7.
In each case given below, find the value of x so that POQ is straight line
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 59

Solution:

Question 8.
in each case, given below, draw perpendicular to AB from an exterior point P
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 64
Solution:

Question 9.
Draw a line segment BC = 8 cm. Using set-squares, draw ∠CBA = 60° and ∠BCA = 75°. Measure the angle BAC. Also measure the lengths of AB and AC.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 68

Question 10.
Draw a line AB = 9 cm. Mark a point P in AB such that AP=5 cm. Through P draw (using set-square) perpendicular PQ = 3 cm. Measure BQ.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 69

Question 11.
Draw a line segment AB = 6 cm. Without using set squares, draw angle OAB = 60° and angle OBA = 90°. Measure angle AOB and write this measurement.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 70

Question 12.
Without using set squares, construct angle ABC = 60° in which AB = BC = 5 cm. Join A and C and measure the length of AC.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 25 Properties of Angles and Lines image - 71

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles

August 11, 2022May 30, 2018 by Prasanna

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles (Including Types, Properties and Constructions)

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles (Including Types, Properties and Constructions)

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

Selina Class 6 Maths ICSE Solutions Physics Chemistry Biology Geography History & Civics

IMPORTANT POINTS

1. Collinear Points: Three or more points which lie on the same straight line, are called collinear points.
2. Non-Collinear Points: Three or more points which do not lie on the same line, are called non- col linear points.
3. Triangle: By joining the three non-collinear points, a triangle is formed or A triangle is a figure which is enclosed by three lines segments. In the figure, ABC is a triangle.
4. Parts of triangle: A triangle has six parts, three sides and three angles which are on the vertices of the triangle.
5. Sum of angles of a triangle: The sum of the three angles of a triangle is 180°.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 1
6. Exterior angle of a triangle: If one side of a triangle is produced then the exterior angle is formed. Exterior angle of a triangle is equal to sum of its interior opposite angles. In other words, we can say that exterior angle of a triangle is greater than each of its interior opposite angles. In the figure.
∠ACE is exterior angle and ∠ACE = ∠A + ∠B and also ∠ACE > ∠A and ∠ACF > ∠B.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 2

7. Classification of triangles :
(A) According to their sides.
(i) Equilateral Triangle: If three sides of a triangle are equal, it is called an equilateral triangle.
(ii) Isosceles Triangle: If any two sides of a triangle are equal, then it is called an isosceles triangle.
(iii) Scalene Triangle: If no two sides of the triangle are equal. Then it is called a scalene triangle.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 3
(B) According to Angles :
(i) Acute-angled Triangle: A triangle whose each angle is acute, is called an acute angled triangle.
(ii) Right-angled Triangle: A triangle whose one angle is a right angle i.e. 90°, is called a right angled triangle.
(iii) Obtused-angled Triangle: A triangle whose one angle is an obtused angle, is called an obtused-angled triangle.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 4

8. Some special properties of a triangle:
(i) If one angle of a triangle is equal to the sum of other two cycles, the angle is a right angled.
(ii) If the acute angles of a right angled triangle are equal, then the triangle is a right angled isosceles triangle and its each acute angle will be of 45°.
(iii) Sum of two sides of a triangle is greater than the third side.
(iv) There can be only one right angle in a triangle.
(v) There can be only one obtuse angle in a triangle.
(vi) Side opposite to greater angle is greater.
(vii) Angle opposite to shorter side is shorter.

Triangles Exercise 26A – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
In each of the following, find the marked unknown angles :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 5

Solution:
(i) Since, sum of all angles of triangle = 180°
Hence, 70° + 72° + z = 180°
⇒ 142°+ z = 180° ”
⇒ z= 180°-142°
z = 38°
(ii) Since, sum of all angles of a triangle = 180°
1st Triangle 50° + 80° + b = 180°
⇒ 130°+ &= 180°
⇒b= 180° – 130°
b = 50°
IInd Triangle 40° + 45° + a = 180°
⇒ 85° + a = 180°
⇒ a = 180° -85
a = 95°
(iii) 60° + 45° + 20° + x = 180°
⇒ 125° + x = 180°
⇒ x = 180° – 125° => x = 55°

Question 2.
Can a triangle together have the following angles ?
(i) 55°, 55° and 80°
(ii) 33°, 74° and 73°
(iii) 85°, 95° and 22°.
Solution:
(i) Sum of all angles of a triangle = 180° Here, 55° + 55° + 80° = 180°
190° ≠ 180°
No.
(ii) 33°+ 74°+ 73°= 180°
180°= 180°
Yes.
(iii) 85° + 95° + 22° = 180°
202° ≠ 180°
No.

Question 3.
Find x, if the angles of a triangle are:
(i) x°, x°, x°
(ii) x°, 2x°, 2x°
(iii) 2x°, 4x°, 6x°
Solution:
(i) Since, sum of all the angles of a triangle =180
x° + x° + x° = 180
⇒ 3x° = 180
⇒ x° = \(\frac { 180 }{ 3 }\)
x = 60
(ii) x° + 2x° + 2x° = 180
5x° = 180
x° = \(\frac { 180 }{ 5 }\)
x° = 36
(iii) 2x° + 4x° + 6x° =180
12x° = 180
x° = \(\frac { 180 }{ 12 }\)
x° = 15

Question 4.
One angle of a right-angled triangle is 70°. Find the other acute angle.
Solution:
We know that, sum of angles of a triangle = 180°.
Let, the acute angle be ‘x’
∴ x + 90° + 70° = 180°
⇒ x+ 160° = 180°
⇒ x= 180°-160°
⇒ x = 20°
∴The acute angle is 20°.

Question 5.
In ∆ABC, ∠A = ∠B = 62° ; find ∠C.
Solution:
∠A + ∠B + ∠C= 180°
⇒ 62° + 62° + ∠C = 180°
⇒ 124° + ∠C = 180°
⇒ ∠C = 180° – 124°
⇒∠C = 56°

Question 6.
In ∆ABC, C = 56°C = 56° ∠B = ∠C and ∠A = 100° ; find ∠B.
Solution:
∠A + ∠B + ∠C = 180°
⇒ 100° + ∠B + ∠B = 180°
⇒ 2∠B = 180° 100°
∠B = \(\frac { 80 }{ 2 }\)°
∠B = 40°
∠C = ∠B = 40°

Question 7.
Find, giving reasons, the unknown marked angles, in each triangle drawn below:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 7

Solution:
We know that,
Exterior angle of a triangle is always equal to the sum of its two interior opposite angles (property)
(i) ∴ 110° = x + 30° (by property)
⇒x=110°-30° x = 80°
(ii) x+115° = 180°
(linear property of angles)
⇒x = 180°- 115° ⇒x = 65°
∴115° = x + y
⇒ 115° = 65° + _y ⇒ y= 115° – 65° =50°
y = 50°
(iii) 110° = 2x + 3x
5x – 110°
x = \(\frac { 110 }{ 5 }\)°
x = 22°
∴2x = 2 x 22 = 44°
3x = 3 x 22 = 66°

Question 8.
Classify the following triangles according to angle :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 9

Solution:
(i) Since, it has an obtuse angle of 120° Hence, it is obtuse angled triangle.
(ii) Since, all the angle of triangle is less than 90°.
Hence, it is an acute angled triangle.
(iii) Since ∠MNL = 90°, and sum of two acute angle ∠M + ∠N = 30° + 60° = 90°.
Hence, it is a right angled triangle.

Question 9.
Classify the following triangles according to sides :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 11
Solution:
(i) Since, two sides-are equal.
Hence, Isosceles triangle.
(ii) Since, all the three sides are unequal.
Hence, Scalene, triangle.
(iii) Since, all the three sides are unequal Hence, Scalene triangle.
(iv) All the three sides are equal.
Hence, equilateral triangle.

Triangles Exercise 26B – Selina Concise Mathematics Class 6 ICSE Solutions

Construct traingle ABC, when :
Question 1.
AB = 6 cm, BC = 8 cm and AC = 4 cm.
Solution:
Steps of Construction:
(1) Draw a line AB = 6 cm.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 12
(2) compasses and taking B as centre, draw an arc of 8 cm radius.
(3) With A as centre, draw an arc of 4 cm radius, which cuts the previous arc at C.
(4) Join AC and BC.
Triangle ABC, obtained, is the required triangle.

Question 2.
AB = 3.5 cm, AC = 4.8 cm and BC = 5.2 cm.
Solution:
Steps of Construction :
(1) Draw a line AB = 3.5 cm.
(2) Using compasses and taking B as centre, draw an arc of 5.2 cm radius.
(3) With A as centre, draw an arc of 4.8
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 13
(4) Join AC and BC

Question 3.
AB = BC = 5 cm and AC = 3 cm. Mea¬sure angles A and C. Is ∠A = ∠C?
Solution:
Steps of Construction :
(1) Draw a line AB = 5 cm.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 14
(2) Using compasses and taking B as cen¬tre, draw an arc of 5 cm radius.
(3) With A as centre, draw an arc of 3 cm radius, which cuts the previous arc at C.
(4) Join AC and BC.

Question 4.
AB = BC = CA = 4.5 cm. Measure all the angles of the triangle. Are they equal ?
Solution:
Steps of Construction :
(1) Draw a a line AB =4.5
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 15
(2) Using compasses and taking BC as centre, draw an arc of 4.5 cm radius.
(3) With AC as centre, draw an arc of 4-5 cm radius, which cuts the previous arc at C.
(4) Join AC and BC.
(5) Measurement, ∠A = ∠B = ∠C = 60°.

Question 5.
AB = 3 cm, BC = 7 cm and ∠B = 90°.
Solution:
Steps of Construction :
(1) Draw a line segment AB = 3 cm.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 16
(2) With the help of compasses, construct ∠ABC = 90°.
(3) With B as centre, draw an arc of 7 cm length which cuts BP at C.
(4) Join A and C.
(5) Triangle ABC, so obtained, is the required triangle.

Question 6.
AC = 4.5 cm, BC = 6 cm and ∠C = 60°.
Solution:
Steps of Construction :
(1) Draw a line AC = 4.5 cm.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 17
(2) With the help of compasses, construct ∠ACB = 60°.
(3) With C as centre, draw an arc of 6 cm radius, which cuts CB at C.
(4) Join B and A.

Question 7.
AC = 6 cm, ∠A = 60“ and ∠C = 45°. Measure AB and BC.
Solution:
Steps of Construction :
(1) Draw a line segment AC = 6 cm.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 18.
(2) At A construct an angle ∠A = 60°.
(3) At C construct an angle ∠C = 45°.
(4) AD and CE intersect each other at B.
(5) ∴ ∆ABC is the required triangle.
(6) On measuring AB = 4-4 cm, BC = 5.4 cm.

Question 8.
AB = 5.4 cm, ∠A = 30° and ∠B = 90°. Measure ∠C and side BC.
Solution:
Steps of Construction :
(1) Draw a line segment AB = 5.4 cm.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 19
(2) At A construct an angle ∠A = 30°.
(3) At B construct an angle ∠B = 90°.
(4) AD and BE intersect each other at C.
(5) ∴ ∆ABC is the required triangle.
(6) On measuring ∠C = 60° side BC = 31 cm.

Question 9.
AB = 7 cm, ∠B = 120° and ∠A = 30°. Measure AC and BC.
Solution:
Steps of Construction :
(1) Draw a line segment AB = 7 cm
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 20
(2) At A construct an angle ∠A = 30°.
(3) At C construct an angle ∠C = 45°.
(4) AE and BD intersect each other at C.
(5) ∴ ∆ABC is the required triangle.
(6) On measuring length of AC = 12cm and BC = 7 cm respectively.

Question 10.
BC = 3 cm, AC = 4 cm and AB = 5 cm. Measure angle ACB. Give a special name to this triangle.
Solution:
Steps of Construction :
(1) Draw a line segment AB = 5 cm
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 21
(2) From A, with the help of compass cut arc AC = 4cm
(3) From point B cut an arc BC = 3 cm.
(4) Join AC and BC.
(5) AABC is required right-angled triangle.
Measuring ∠ACB = 90°

Triangles Revision Exercise – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
If each of the two equal angles of an isosceles triangle is 68°, find the third angle.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 22

Question 2.
One of the angles of a triangle is 110°, the two other angles are equal. Find their value.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 23

Question 3.
The angles of a triangle are in the ratio 3:5: 7. Find each angle.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 25

Question 4.
The angles of a triangle are (2x – 30°),(3x – 40°) and (\(\frac { 5 }{ 2 }\)x + 10°) Find the value of x .
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 26

Question 5.
In each of the following figures, triangle ABC is equilateral and triangle PBC is isosceles. If PBA = 20°; find in each case:
(a) angle PBC.
(b) angle BPC.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 28
Solution:

Question 6.
Construct a triangle ABC given AB = 6 cm, BC = 5 cm and CA = 5.6 cm. From vertex A draw a perpendicular on to side BC. Measure the length of this perpendicular.
Solution:
Steps of Construction :
(1) Draw a line AB = 6 cm.
(2) Using compass, taking A and B as centre draw arcs of 5 cm and 5.6 cm respectively, which cut each other at C.
(3) Join AC and BC.
(4) Now, from vertex A draw a bisector AD towards BC.
On measuring length AD = 5 cm.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 31

Question 7.
Construct a triangle PQR, given PQ = 6 cm, ∠P = 60° and ∠Q = 30°. Measure angle R and the length of PR.
Solution:
Steps of Construction :
(1) Draw a line PQ = 6 cm.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 32
(2) Using compass taking P as centre draw an angle ∠P = 60°.
(3) Using compass taking Q as a centre draw an angle ∠Q = 30°.
(4) PS and QT intersect each other R.
(5) ∆RPQ is the required triangle.
On measuring; ∠R = 90°, length of PR = 3 cm.

Question 8.
Construct a triangle ABC given BC = 5 cm, AC = 6 cm and ∠C = 75°. Draw the bisector of the interior angle at A. Let this bisector meet BC at P ; measure BP.
Solution:
Steps of Construction :
(1) Draw BC = 5 cm.
(2) With the help of compass from centre C. Draw an angle ∠C = 75°.
(3) From CD, cut an arc AC = 6 cm.

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 33
(4) JoinAB.
(5) From A draw an bisector AP.
(6) On measuring BP = 2.6 cm.

Question 9.
Using ruler and a pair compass only, construct a triangle XYZ given YZ = 7 cm, ∠XYZ = 60° and ∠XZY = 45°. Draw the bisectors of angles X and Y.
Solution:
Steps of Construction :
(1) Draw a line YZ = 7 cm.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 34
(2) Y as a centre draw an ∠XYZ = 60°.
(3) Z as a centre draw an ∠XZY = 45°.
(4) From X and Y as centre draw bisector of ∠X and ∠Y, which meet at point O.

Question 10.
Using ruler and a pair compass only, construct a triangle PQR, given PQ = 5.5 cm, QR = 7.5 cm and RP = 6 cm. Draw the bisectors of the interior angles at P, Q and R. Do these bisectors meet at the same point ?
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 35

Steps of Construction :
(1) Draw a line PQ = 5.5 cm.
(2) From Q as a centre draw an arc QR = 7.5 cm.
(3) From P as a centre draw an arc PR = 6 m, which intersects previous arc at R.
(4) Join PR and QR.
(5) Now, draw interior bisectors of ∠P, ∠QR ∠R which meets each other at point S.

Question 11.
One angle of a triangle is 80° and the other two are in the ratio 3 : 2. Find the unknown angles of the triangle.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 36
Let angle ∠A of a triangle ABC = 80°
But sum of three angles of a triangle = 180°
Sum of remaining two angles = 180° – 80° = 100°
Ratio in their two angles = 3:2
Let second angle = 3x
and third angle = 2x
3x + 2x = 100°
5x = 100 x = \(\frac { 100 }{ 5 }\) = 20°
second angle ∠B = 3x = 3 x 20° = 60°
and third angle ∠C = 2x = 2 x 20° = 40°
Hence other two angles are 60° and 40°.

Question 12.
Find the value of x if ∠A = 32°, ∠B = 55° and obtuse angle AED = 115°.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 37
In the figure, ∠A = 32, ∠B = 55°
∠AED =115°
In ∆ABC
Exterior ∠ACD = ∠A + ∠B = 32° + 55° = 87°
Similarly in ∆CDE
Ext. ∠AED = ∠D + ∠ACD
⇒ 115° = x + 87° ⇒ x = 115°- 87° = 28°
Hence x = 28°

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral

August 11, 2022May 30, 2018 by Prasanna

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral

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IMPORTANT POINTS
4.Quadrilateral: A quadrilateral is a plane figure enclosed by four sides. It has four sides, four interior angles and four vertices.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 1
In quadrilateral ABCD, shown alongside:
(i) four sides are : AB, BC, CD and DA.
(ii) four angles are : ∠ABC,∠BCD, ∠CDA and ∠DAB ; which are numbered∠1, ∠2, ∠3 and ∠4 respectively.
(iii) four vertices are : A, B, C and D.

5. Diagonals of a Quadrilateral : The line segments joining the opposite vertices of a quadrilateral are called its diagonals.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 2
The given figure shows a quadrilateral PQRS with diagonals PR and QS.
6. Types of Quadrilaterals :
1. Trapezium: A trapezium is a quadrilateral in which one pair of opposite sides are parallel.

The figure, given alongside, shows a trapezium as its sides AB and DC are parallel i.e. AB || DC.
When the non-parallel sides of the trapezium are equal in length, it is called an isosceles trapezium.
The given figure shows a trapezium ABCD whose non-parallel sides AD and BC are equal in length i.e. AD = BC; therefore it is an isosceles trapezium.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 4
Also, in an isosceles trapezium :

(i) base angles are equal:
i.e. ∠A = ∠B and ∠D =∠C
(ii) diagonals are equal
i.e. AC = BD.

2.Parallelogram : A parallelogram is a quadrilateral, in which both the pairs of opposite sides are parallel.
The quadrilateral ABCD, drawn alongside, is a parallelogram; since, AB is parallel to DC and AD is parallel to BC i.e.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 6
AB || DC and AD || BC.
Also, in a parallelogram ABCD:
(i) opposite sides are equal:
i.e. AB = DC and AD = BC.
(ii) opposite angles are equal:
i.e. ∠ABC = ∠ADC and ∠BCD = ∠BAD
(iii) diagonals bisect each other :
i.e. OA = OC = \(\frac { 1 }{ 2 }\) AC and OB = OD = \(\frac { 1 }{ 2 }\) BD.

7. Some special types of Parallelograms
(a) Rhombus : A rhombus is a parallelogram in which all its sides are equal.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 9
∴In a rhombus ABCD :
(i) opposite sides are parallel:
i.e. AB||DC and AD||BC.
(ii) all the sides are equal:
i.e. AB = BC = CD = DA.
(iii) opposite angles are equal:
i.e. ∠A = ∠C and ∠B = ∠D.
(iv) diagonals bisect each other at right angle :
i.e. OA= OC = \(\frac { 1 }{ 2 }\) AC ; OB = OD = \(\frac { 1 }{ 2 }\) BD.
and ∠AOB= ∠BOC = ∠COD = ∠AOD = 90°
(v) diagonals bisect the angles at the vertices :
i.e. ∠1 =∠2 ;∠3 = ∠4 ; ∠5 = ∠6 and ∠7 =∠8.

(b) Rectangle : A rectangle is a parallelogram whose any angle is 90°.
A rectangle is also defined as a quadrilateral whose each angle is 90°.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 10
Note : If any angle of a parallelogram is 90° ; automatically its each angle is 90° ; the reason being that the opposite angles of a parallelogram are equal.
Also, in a rectangle:
(i) opposite sides are parallel.
(ii) opposite sides are equal.
(iii) each angle is 90°.
(iv) diagonals are equal.
(v) diagonals bisect each other.

(c) Square : A square is a parallelogram, whose all side are equal and each angle is 90°.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 11
A square can also be defined as :
(i) a rhombus whose any angle is 90°.
(ii) a rectangle whose all sides are equal.
(iii) a quadrilateral whose all sides are equal and each angle is 90°.
∴ If ABCD is a square :
(i) all its sides are equal, i.e. AB = BC = CD = DA
(ii) each angle of it is 90°.
i.e. ∠A = ∠B = ∠C = ∠D = 90°.
Also,
(iii) diagonals are equal.
i.e. AC = BD.
(iv) diagonals bisect each other at 90°.
i.e. OA = OC =\(\frac { 1 }{ 2 }\) AC;OB = OD = \(\frac { 1 }{ 2 }\) BD
and ∠AOB = ∠BOC = ∠COD = ∠DOA = 90°.
Since, diagonals AC and BD are equal; therefore ; OA = OC = OB = OD.
(v) diagonals bisect the angles at the vertices
i.e. ∠1 = ∠2 = 45° [∵ ∠1 + ∠2 = 90°]
Similarly; ∠3 = ∠4 = 45° ;
∠5 – ∠6 = 45° and ∠7 = ∠8 = 45°.

Quadrilateral Exercise 27A – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Two angles of a quadrilateral are 89° and 113°. If the other two angles are equal; find the equal angles.
Solution:
Let the other angle = x°
According to given,
89° + 113° + x° + x° = 360°
2x° = 360° – 202°
2x° = 158°
x° = \(\frac { 158 }{ 2 }\) =79°
∴other two angles = 79° each

Question 2.
Two angles of a quadrilateral are 68° and 76°. If the other two angles are in the ratio 5 : 7; find the measure of each of them.
Solution:
Two angles are 68° and 76°
Let other two angles be 5x and 7x
∴ 68°+76°+5x + 7x = 360°
12x + 144° = 360°
12x = 360° – 144°
12x= 216°
x = 18°
angles are 5x and 7x
i.e. 5×18° and 7×18° i.e. 90° and 126°

Question 3.
Angles of a quadrilateral are (4x)°, 5(x+2)°, (7x-20)° and 6(x+3)°. Find
(i) the value of x.
(ii) each angle of the quadrilateral.
Solution:
Angles of quadrilateral are,
(4x)°, 5(x+2)°, (7x-20)° and 6(x+3)°.
4x+5(x+2)+(7x-20)+6(x+3) = 360°
4x+5x+10+7x-20+6x+18 = 360° 22x+8 = 360°
22x = 360°-8°
22x = 352°
x = 16°
Hence angles are,
(4x)° = (4×16)° = 64°,
5(x+2)° = 5(16+2)° = 90°,
(7x-20)° = (7×16-20)° = 92°
6(x+3)° = 6(16+3) = 114°

Question 4.
Use the information given in the following figure to find :
(i) x
(ii) ∠B and ∠C
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 12
Solution:
∵ ∠A = 90° (Given)
∠B = (2x+4°)
∠C = (3x-5°)
∠D = (8x – 15°)
∠A + ∠B + ∠C + ∠D = 360°
90° + (2x + 4°) + (3x – 5°) + (8x – 15°) = 360°
90° + 2x + 4° + 3x – 5° + 8x – 15° = 360°
⇒ 74° + 13x = 360°
⇒13x = 360° – 74°
⇒ 13x = 286°
⇒ x = 22°
∵ ∠B = 2x + 4 = 2. x 22° + 4 – 48°
∠C = 3x – 5 = 3×22° -5 = 61°
Hence (i) 22° (ii) ∠B = 48°, ∠C = 61°

Question 5.
In quadrilateral ABCD, side AB is parallel to side DC. If ∠A : ∠D = 1 : 2 and ∠C : ∠B = 4:5
(i) Calculate each angle of the quadrilateral.
(ii) Assign a special name to quadrilateral ABCD.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 13
∵∠A : ∠D = 1:2
Let ∠A = x and ∠B = 2x
∵∠C : ∠B = 4 : 5 Let ∠C = 4y and ∠B = 5y
∵AB || DC
∠A + ∠D = 180° x + 2x = 180°
3x = 180° x = 60°
∴A = 60°
∠D = 2x = 2 x 60 = 120° Again ∠B + ∠C = 180°
5y + 4y= 180°
9y = 180°
y = 20°
∴∠B = 5y- = 5 x 20 = 100°
∠C = 4y = 4 x 20 = 80°
Hence ∠A = 60° ; ∠B = 100° ; ∠C = 80° and ∠D = 120°

Question 6.
From the following figure find ;
(i) x,
(ii) ∠ABC,
(iii) ∠ACD.
Solution:
(i) In Quadrilateral ABCD,
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 14
x + 4x + 3x + 4x + 48° = 360°
12x = 360° – 48°
12x =312
(ii) ∠ABC = 4x
4 x 26 = 104°
(iii) ∠ACD = 180°-4x-48°
= 180°-4×26°-48°
= 180°-104°-48°
= 180°-152° = 28°

Question 7.
Given : In quadrilateral ABCD ; ∠C = 64°, ∠D = ∠C – 8° ;
∠A = 5(a+2)° and ∠B = 2(2a+7)°.
Calculate ∠A.
Solution:
∵∠C = 64° (Given)
∴∠D = ∠C- 8°
= 64°- 8°
= 56°
∠A = 5 (a + 2)°
∠B = 2(2a + 7)°
Now ∠A + ∠B + ∠C + ∠D = 360°
5(a + 2)° + 2(2a + 7)° + 64°+ 56° = 360°
5a + 10 + 4a + 14° + 64° + 56° = 360°
9a + 144° = 360°
9a = 360°-144°
9a = 216°
a = 24°
∴∠A = 5(a + 2)
= 5(24 + 2)
= 130°

Question 8.
In the given figure :
∠b = 2a + 15
and ∠c = 3a+5; find the values of b and c.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 15
Solution:
∠b = 2a + 15
& ∠c = 3a + 5
∵Sum of angles of quadrilateral = 360°
70° + a + 2a + 15 + 3a + 5 – 360°
6a+90° = 360°
6a = 270°
a = 45°
∴ b = 2a+15= 2×45+15 = 105°
c = 3a+5 = 3×45+5 = 140°
105° and 140°

Question 9.
Three angles of a quadrilateral are equal. If the fourth angle is 69°; find the measure of equal angles.
Solution:
Let each equal angle be
x° x + x + x + 69° = 360°
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 16
3x = 360°-69 3x =291 x = 97°
Each equal angle = 97°

Question 10.
In quadrilateral PQRS, ∠P : ∠Q : ∠R : ∠S = 3 : 4 : 6 : 7.
Calculate each angle of the quadrilateral and then prove that PQ and SR are parallel to each other. Is PS also parallel to QR ?
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 17
∵∠P : ∠Q : ∠R : ∠S = 3 : 4 : 6 : 7
Let ∠P = 3x
∠Q = 4x
∠R = 6x & ∠S = 7x
∴∠P+∠Q+∠R+∠S = 360°
3x + 4x + 6x + lx = 360°
20x = 360°
x = 18°
∴ ∠P = 3x = 3×18 = 54°
∠Q = 4x = 4×18 = 72°
∠R = 6x = 6×18 = 108°
∠S = 7x = 7×18 = 126°
∠Q + ∠R = 72°+108° = 180° or ∠P +∠S = 54°+126° = 180°
Hence PQ || RS
As ∠P + ∠Q = 72° +54° = 126°
Which is * 180°.
∴PS and QR are not parallel.

Question 11.
Use the information given in the following figure to find the value of x.
Solution:
Take A, B, C, D as the vertices of quadrilateral and BA is produced to E (say).
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 18
Since ∠EAD = 70°
∴∠DAB = 180° – 70° = 110° [∵ EAB is a straight line and AD stands on it]
∴∠EAD + ∠DAB = 180°
∴110° + 80° + 56° + 3x = 360°
[∵ sum of interior angles of a quadrilateral = 360°]
∴3x = 360° – 110° – 80° – 56° + 6°
3x = 360° – 240° = 120°
∴x = 40°

Question 12.
The following figure shows a quadrilateral in which sides AB and DC are parallel.
If ∠A : ∠D = 4 : 5, ∠B = (3x – 15)° and ∠C = (4x + 20)°, find each angle of the quadrilateral ABCD.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 19
Solution:
Let ∠A = 4x
∠D = 5x
Since ∠A + ∠D = 180° [∵ AB || DC]
∴4x + 5x = 180°
⇒ 9x = 180° ⇒x = 20°
∴∠A = 4 (20) = 80°, ∠D = 5 (20) = 100° Again ∠B + ∠C = 180° [ ∵ AB || DC]
∴ 3x – 15° + 4x + 20° = 180°
7x = 180°-5°
⇒ 7x = 175° ⇒ x = 25°
∴∠B = 75°-15° = 60° and ∠C = 4(25) + 20 = 100° + 20° = 120°

Quadrilateral Exercise 27B – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
In a trapezium ABCD, side AB is parallel to side DC. If ∠A = 78° and ∠C = 120°, find angles B and D.
Solution:
∵ AB || DC and BC is transversal
∴∠B and ∠C, ∠A and ∠D are Cointerior angles with their sum = 180°
i.e. ∠B + ∠C = 180°
⇒ ∠B + 120° = 180°
⇒ ∠B = 180° – 120°
⇒ ∠B = 60°
Also ∠A + ∠D = 180°
⇒ 78° + ∠D = 180°
⇒ ∠D = 180° – 78°
∠D = 102°

Question 2.
In a trapezium ABCD, side AB is parallel to side DC. If ∠A = x° and ∠D = (3x – 20)°; find the value of x.
Solution:
∵AB || DC and BC is transversal
∴∠A and ∠B are Co-interior angles with their sum = 180°
i.e. ∠A + ∠D = 180°
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 20
⇒ x° + (3x – 20)° = 180°
⇒ x° + 3x° – 20° = 180°
⇒ 4x° = 180° + 20°
x° = \(\frac { 200 }{ 4 }\) = 50°
∴Value of x = 50°

Question 3.
The angles A, B, C and D of a trapezium ABCD are in the ratio 3 : 4 : 5 : 6.
Le. ∠A : ∠B : ∠C : ∠D = 3:4: 5 : 6. Find all the angles of the trapezium. Also, name the two sides of this trapezium which are parallel to each other. Give reason for your answer
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 21

Question 4.
In an isosceles trapezium one pair of opposite sides are ….. to each Other and the other pair of opposite sides are ….. to each other.
Solution:
In an isosceles trapezium one pair of opposite sides are parallel to each other and the other pair of opposite sides are equal to each other.

Question 5.
Two diagonals of an isosceles trapezium are x cm and (3x – 8) cm. Find the value of x.
Solution:
∵The diagonals of an isosceles trapezium are of equal length
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 23
∴3x – 8 = x
⇒ 3x – x = 8 cm
⇒ 2x = 8 cm
⇒ x = 4 cm
∴ The value of x is 4 cm

Question 6.
Angle A of an isosceles trapezium is 115° ; find the angles B, C and D.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 24

Question 7.
Two opposite angles of a parallelogram are 100° each. Find each of the other two opposite angles.
Solution:
Given : Two opposite angles of a parallelogram are 100° each
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 25
∵ Adjacent angles of a parallelogram are supplementary,
∴∠A + ∠B = 180°
⇒ 100° + ∠B = 180°
⇒ ∠B = 180° – 100°
⇒ ∠B = 80°
Also, opposite angles of a parallelogram are equal
∴∠D = ∠B = 80°
∴∠B = ∠D = 80°

Question 8.
Two adjacent angles of a parallelogram are 70° and 110° respectively. Find the other two angles of it.
Solution:
Given two adjacent angles of a parallelogram are 70° and 110° respectively.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 26
Since, we know that opposite angles of a parallelogram are equal
∴∠C = ∠A = 70° and ∠D = ∠B = 110°

Question 9.
The angles A, B, C and D of a quadrilateral are in the ratio 2:3: 2 : 3. Show this quadrilateral is a parallelogram.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 27

Question 10.
In a parallelogram ABCD, its diagonals AC and BD intersect each other at point O.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 29
If AC = 12 cm and BD = 9 cm ; find; lengths of OA and OD.
Solution:

Question 11.
In parallelogram ABCD, its diagonals intersect at point O. If OA = 6 cm and OB = 7.5 cm, find the length of AC and BD.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 32
Solution:

Question 12.
In parallelogram ABCD, ∠A = 90°
(i) What is the measure of angle B.
(ii) Write the special name of the paralleogram.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 34

Question 13.
One diagnol of a rectangle is 18 cm. What is the length of its other diagnol?
Solution:
∵ In a rectangle, diagnols are equal
⇒ AC = BD
Given, one diagnol of a rectangle = 18cm
∴ Other diagnol of a rectangle will be = 18cm
i.e. AC = BD = 18cm.

Question 14.
Each angle of a quadrilateral is x + 5°. Find :
(i) the value of x
(ii) each angle of the quadrilateral.
Give the special name of the quadrilateral taken.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 36
Solution:

Question 15.
If three angles of a quadrilateral are 90° each, show that the given quadrilateral is a rectangle.
Solution:
The given quadrilateral ABCD will be a rectangle, if its each angle is 90°
Since, the sum of interior angles of a quadrilateral is 360°.
∴∠A +∠B + ∠C + ∠D = 360°
⇒ 90° + 90° + 90° + ∠D = 360°
⇒ 270° + ∠D = 360°
⇒ ∠D = 360° – 270°
⇒ ∠D = 90°
Since, each angle of the quadrilateral is 90°.
∴The given quadrilateral is a rectangle.

Question 16.
The diagnols of a rhombus are 6 .cm and 8 cm. State the angle at which these diagnols intersect.
Solution:
The diagnols of a Rhombus always intersect at 90°.

Question 17.
Write, giving reason, the name of the figure drawn alongside. Under what condition will this figure be a square.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 39
Solution:
Since, all the sides of the given figure are equal.
i.e. AB = BC = CD = DA = 6 cm
∴ The given figure is a rhombus.
This figure shall be considered as a square, if any angle is 90°.

Question 18.
Write two conditions that will make the adjoining figure a square.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 40
Solution:
The conditions that will make the ad-joining figure a square are :
(i) All the sides must be equal.
(ii) Any angle is 90°.

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons

August 11, 2022May 29, 2018 by Prasanna

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

Selina Class 6 Maths ICSE Solutions Physics Chemistry Biology Geography History & Civics

IMPORTANT POINTS

1. Polygon : A closed plane geometrical figure, bounded by atleast three line segments, is called a
polygon.
The adjoining figure is a polygon as it is :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 1
(i) Closed
(ii) bounded by five line segments AB, BC, CD, DE and AE.
Also, it is clear from the given polygon that:
(i) the line segments AB, BC, CD, DE and AE intersect at their end points.
(ii) two line segments, with a common vertex, are not collinear i.e. the angle at any vertex is not 180°.
A polygon is named according to the number of sides (line-segments) in it:

Note : No. of sides :	3	4	5	6
Name of polygon :	Triangle	Quadrilateral	Pentagon	Hexagon

2. Sum of Interior Angles of a Polygon
1. Triangle : Students already know that the sum of interior angles of a triangle is always 180°.
i.e. for ∆ ABC, ∠B AC + ∠ABC + ∠ACB = 180°
⇒ ZA + ZB + ZC = 180°
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 2

2. Quadrilateral : Consider a quadrilateral ABCD as shown alongside.
If diagonal AC of the quadrilaterals drawn, the quadrilateral will be divided into two triangles ABC and ADC.
Since, the sum of interior angles of a triangle is 180°.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 3
∴ In ∆ ABC, ∠ABC + ∠BAC +∠ACB = 180°
And, in ∆ ADC ∠DAC + ∠ADC + ∠ACD = 180°
Adding we get:
∠ABC + ∠BAC +∠ACB + ∠DAC + ∠ADC + ∠ACD = 180° +180°
⇒(∠BAC + ∠DAC) + ∠ABC + (∠ACB + ∠ACD) + ∠ADC = 360°
⇒∠BAD + ∠ABC + ∠BCD + ∠ADC = 360°
⇒ ∠A + ∠B + ∠C + ∠D = 360°
Alternative method : On drawing the diagonal AC, the given quadrilateral is divided into two triangles. And, we know the sum of the interior angles of a triangle is 180°.
∴ Sum of interior angles of the quadrilateral ABCD
= Sum of interior angles of ∆ ABC + sum of interior angles of ∆ ADC = 180°+ 180° = 360°

3. Pentagon : Consider a pentagon ABCDE as shown alongside.
On joining CA and CE, the given pentagon is divided into three triangles ABC, CDE and ACE.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 4
Since, the sum of the interior angles of a triangle is 180°
Sum of the interior angles of the pentagon ABCDE = Sum of interior angles of (∆ ABC + ∆ CDE + ∆ACE)
= 180° + 180° + 180° = 540°

4. Hexagon :
It is clear from the given figure that the sum of the interior angles of the hexagon ABCDEF.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 5
= Sum of inteior angles of
(∆ABC + ∆ ACF + ∆ FCE + ∆ ECD)
= 180° + 180° + 180° + 180° = 720°

3. Using Formula : The sum of interior angles of a polygon can also be obtained by using the following formula:
Note : Sum of interior angles of a polygon = (n – 2) x 180°
where, n = number of sides of the polygon.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 6

Polygons Exercise 28A – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
State, which of the following are polygons :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 7
Solution:
Only figure (ii) and (iii) are polygons.

Question 2.
Find the sum of interior angles of a polygon with :
(i) 9 sides
(ii) 13 sides
(iii) 16 sides
Solution:
(i) 9 sides
No. of sides n = 9
∴Sum of interior angles of polygon = (2n – 4) x 90°
= (2 x 9 – 4) x 90°
= 14 x 90°= 1260°
(ii) 13 sides
No. of sides n = 13
∴ Sum of interior angles of polygon = (2n – 4) x 90° = (2 x 13 – 4) x 90° = 1980°
(iii) 16 sides
No. of sides n = 16
∴ Sum of interior angles of polygon = (2n – 4) x 90°
= (2 x 16 – 4) x 90°
= (32 – 4) x 90° = 28 x 90°
= 2520°

Question 3.
Find the number of sides of a polygon, if the sum of its interior angles is :
(i) 1440°
(ii) 1620°
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 8

Question 4.
Is it possible to have a polygon, whose sum of interior angles is 1030°.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 10

Question 5.
(i) If all the angles of a hexagon arc equal, find the measure of each angle.
(ii) If all the angles of an octagon are equal, find the measure of each angle,
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 12

Question 6.
One angle of a quadrilateral is 90° and all other angles are equal ; find each equal angle.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 13

Question 7.
If angles of quadrilateral are in the ratio 4 : 5 : 3 : 6 ; find each angle of the quadrilateral.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 14

Question 8.
If one angle of a pentagon is 120° and each of the remaining four angles is x°, find the magnitude of x.
Solution:
One angle of a pentagon = 120°
Let remaining four angles be x, x, x and x
Their sum = 4x + 120°
But sum of all the interior angles of a pentagon = (2n – 4) x 90°
= (2 x 5 – 4) x 90° = 540°
= 3 x 180° = 540°
∴ 4x+120o° = 540°
4x = 540° – 120°
4x = 420
x = \(\frac { 420 }{ 4 }\) ⇒ x = 105°
∴Equal angles are 105° (Each)

Question 9.
The angles of a pentagon are in the ratio 5 : 4 : 5 : 7 : 6 ; find each angle of the pentagon.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 15

Question 10.
Two angles of a hexagon are 90° and 110°. If the remaining four angles arc equal, find each equal angle.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 16

Polygons Exercise 28B – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Fill in the blanks :
In case of regular polygon, with
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 17
Solution:

Question 2.
Find the number of sides in a regular polygon, if its each interior angle is :
(i) 160°
(ii) 150°
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 21

Question 3.
Find number of sides in a regular polygon, if its each exterior angle is :
(i) 30°
(ii) 36°
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 22

Question 4.
Is it possible to have a regular polygon whose each interior angle is :
(i) 135°
(ii) 155°
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 24

Question 5.
Is it possible to have a regular polygon whose each exterior angle is :
(i) 100°
(ii) 36°
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 26

Question 6.
The ratio between the interior angle and the exterior angle of a regular polygon is 2 : 1. Find :
(i) each exterior angle of this polygon.
(ii) number of sides in the polygon.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 28

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle

August 11, 2022May 29, 2018 by Prasanna

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

Selina Class 6 Maths ICSE Solutions Physics Chemistry Biology Geography History & Civics

IMPORTANT POINTS

1. A circle is a round enclosed figure, whose mid-point is called its centre.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -1
2. The line segment joining the centre to any point on the circle is called a radius. A centre has infinite radii and all radii of a circle are equal.
3. A line segment which contains the centre of the circle and whose ends points lie on the circle is called diameter of the circle. Diameters of a circle are also equal.
4. Parts of a circle: A circle has three parts (i) Interior (ii) Exterior and (iii) Circle itself.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -2
5. Concentric circles: Two or more circles having the same centre but different radii are called concentric circles.
6. Chord of a circle: A line which divides the circle into two parts is called chord of the circle. Diameter is the longest chord of the circle.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -3
7. Segment of a circle: When a chord divides the circle into two unequal parts, the bigger part is called the major segment and smaller part is called the minor segment.
8. Arc of a circle: A part of circumference of a circle is called an arc of the circle. Arc greater than half circle is called the major arc and less than half circle is called the minor arc.
9. Sector of a circle: A diameter divides the circle into two equal parts and each part is called a semicircle. Sector greater than a semi-circle is called the major sector and less than semi-circle is called the minor sector of the circle.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -4

The Circle Exercise 29A – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Use the figure given below to fill in the blanks :
(i) R is the …… of the circle.
(ii) Diameter of a circle is …… .
(iii) Tangent to a circle is … .
(iv) EF is a …… of the circle.
(v) …… is a chord of the circle.
(vi) Diameter = 2 x …..
(vii) ……. is a radius of the circle.
(viii) If the length of RS is 5 cm, the length of PQ = ……
(ix) If PQ is 8 cm long, the length of RS =…..
(x) AB is a ….. of the circle
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -5
Solution:
(i) center
(ii) PQ
(iii)AB
(iv) secant
(v) CD
(vi) radius
(vii) RS
(viii) 10 cm
(ix) 4 cm
(x) tangent.

Question 2.
Draw a circle of radius 4.2 cm. Mark its centre as O. Take a point A on the circumference of the circle. Join AO and extend it till it meets point B on the circumference of the circle,
(i) Measure the length of AB.
(ii) Assign a special name to AB.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -6
(i) By measurement AB = 8.4 cm.
(ii) ∴ AB is the diameter of the circle.

Question 3.
Draw circle with diameter :
(i) 6 cm
(ii) 8.4 cm.
In each case, measure the length of the radius of the circle drawn.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -7

Question 4.
Draw a circle of radius 6 cm. In the circle, draw a chord AB = 6 cm.
(i) If O is the centre of the circle, join OA and OB.
(ii) Assign a special name to ∆AOB
(iii) Write the measure of angle AOB.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -8

Question 5.
Draw a circle of radius 4.8 cm and mark its centre as P.
(i) Draw radii PA and PB such that ∠APB = 45°.
(ii) Shade the major sector of the circle
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -9
PA is the radius of circle. i.c., PA = 4.8 cm.
(i) ∠APB = 45° in which P is the centre of the circle and PA and PB are radii of circle.
(ii) Major sector of circle is shaded in the figure.

Question 6.
Draw a circle of radius 3.6 cm. In the circle, draw a chord AB = 5 cm. Now shade the minor segment of the circle.
Solution:
Shaded portion of circle is the minor segment of the circle.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -10

Question 7.
Mark two points A and B ,4cm a part, Draw a circle passing through B and with A as a center
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -11
Solution:
In the figure, A is the centre of the circle and AB = 4 cm [radius of circle]

Question 8.
Draw a line AB = 8.4 cm. Now draw a circle with AB as diameter. Mark a point C on the circumference of the circle. Measure angle ACB.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -12
By measurement ∠ACB =90

The Circle Exercise 29B – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Construct a triangle ABC with AB = 4.2 cm, BC = 6 cm and AC = 5cm. Construct the circumcircle of the triangle drawn.
Solution:
Steps of Construction :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -13
(i) Draw ∆ABC in which AB = 4.2 cm. BC = 6 cm. and AC = 5 cm.
(ii) Draw the perpendicular bisectors of any two sides of the triangle. Let these intersect at O.
(iii) Taking O as centre and OA or OB or OC as radius draw a circle.
This circle will pass through vertices A, B and C.

Question 2.
Construct a triangle PQR with QR = 5.5 cm, ∠Q = 60° and angle R = 45°.
Construct the circumcircle cif the triangle PQR.
Solution:
Steps of Construction :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -14
(i) Draw a ∆PQR in which QR = 5.5 cm, ∠Q = 60° and ∠R = 45°.
(ii) Draw the arc bisector of PQ and PR which intersect at O.
(iii) Taking O as centre and radius OP or OQ or OR draw a circle.
This circle will pass through vertices P, Q and R.

Question 3.
Construct a triangle ABC with AB = 5 cm, ∠B = 60° and BC = 6. 4 cm.
Draw the incircle of the triangle ABC. Sol. Steps of Construction :
Solution:
Steps of Construction:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -15
(i) Draw a line AB = 5 cm.
(ii)B as a centre draw an angle with the help of compass ∠B = 60°. Cut the line with an arc BC = 6 4 cm.
(iii) Join AC.
(iv) Now, from A and B cut the bisector of ∠A and ∠B, which intersect each other at point D.
(v) With D as a centre draw an incircle which touches all the three sides of AABC.

Question 4.
Construct a triangle XYZ in which XY = YZ= 4.5 cm and ZX = 5.4 cm. Draw the circumcircle of the triangle and measure its circumradius.
Solution:
Steps of Construction :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -16
(i) Draw a triangle XYZ in which XY = YZ = 4.5 cm and ZX = 5.4 cm.
(ii) Draw the bisectors of XZ and YZ which meet at O.
(iii) With O as centre and radius OX or OY or OZ draw a circle.
This circle will pass through X, Y and Z.

Question 5.
Construct a triangle PQR in which, PQ = QR = RP = 5.7 cm. Draw the incircle of the triangle and measure its radius.
Solution:
Steps of Construction :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -17
(i) Draw an equilateral ∆ RPQ in which PQ = QR = RP = 5.7 cm each.
(ii) From P and Q cut the bisector of ∠P and ∠Q, which intersect each other at point O.
(iii) With P as a centre draw an incircle which touches all the three sides of ∆RPQ.

The Circle Revision Exercise – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
The centre of a circle is at point O and its radius is 8 cm. State the position of a point P (point P may lie inside the circle, on the circumference of the circle, or outside the circle), when :
(a) OP = 10.6 cm
(b) OP = 6.8 cm
(c) OP = 8 cm
Solution:
(a) Draw circle each of radius 8 cm. With centre O
In figure (i) draw OP = 10.6 cm
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -18
We see that point P lies outside the circle as OP > radius of the circle
(b) In figure (ii) OP = 6.8 cm. We see that P lies inside the circle as OP < radius of the circle.
(c) In figure, OP = 8 cm. We see that P lies on the circle as OP = radius of the circle.

Question 2.
The diameter of a circle is 12.6 cm. State, the length of its radius.
Solution:
Diameter of the circle = 12.6 cm
∴Radius = \(\frac { 1 }{ 2 }\) diameter = \(\frac { 1 }{ 2 }\) x 12.6 cm
= 6.3 cm

Question 3.
Can the length of a chord of a circle be greater than its diameter ? Explain.
Solution:
No, the length of chord cannot be greater than the diameter of the cirlce as the diameter of a circle is the greatest chord of that circle.

Question 4.
Draw a circle of diameter 7 cm. Draw two radii of this circle such that the angle between these radii is 90°. Shade the minor sector obtained. Write a special name for this sector.
Solution:
Draw a circle with diameter = 7 cm
OA and OB are the radii of the circle such that ∠AOB = 90°
Now shade the minor sector AOB This is the quadrant of the circle
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -19

Question 5.
State, which of following statements are true and which are false :
(i) If the end points A and B of the line segment lie on the circumference of a circle, AB is a diameter.
(ii) The longest chord of a circle is its diameter.
(iii) Every diameter bisects a circle and each part of the circle so obtained is a semi-circle.
(iv) The diameters of a circle always pass through the same point in the circle.
Solution:
(i) False, as AB may be diameter or may not be, it can be chord.
(ii) True, diameter of a circle is the longest chord.
(iii) True.
(iv) True, all the diameter of a circle pass through the same point i.e., centre, of the circle.

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 30 Revision Exercise Symmetry

August 11, 2022May 29, 2018 by Prasanna

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 30 Revision Exercise Symmetry (Including Constructions on Symmetry)

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 30 Revision Exercise Symmetry (Including Constructions on Symmetry)

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

Selina Class 6 Maths ICSE Solutions Physics Chemistry Biology Geography History & Civics

IMPORTANT POINTS

1. Concept of Symmetry (Linear Symmetry) : Consider a plane mirror mm’. If your face F is at a distance ‘d’ before the mirror ; the image F, of your face is also formed behind the mirror at distance “d” from it.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 30 Revision Exercise Symmetry image - 1
Now, if the Figure 1, given alongside, is folded about the mirror mm’, the object F and its image F’ coincide. Since on folding the figure about mm’ its two parts exactly coincide, we say that the figure is symmetrical about the mirror line mm’. For the reason, the mirror line mm’ is called line of symmetry of the whole figure including face F, its image F’ and the mirror mm’.
Keep in Mind :
in order to check, whether a given figure is symmetrical about a line in it or not; fold the figure about that line. If part of the figure, which lies on one side of the line exactly coincides with the part of the figure on the other side of the line, then the figure is symmetrical about that line.
For Example :
(i) A line segment is symmetrical about its perpendicular bisector
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 30 Revision Exercise Symmetry image - 2
(ii) An angle (with equal arms) is symmetrical about its bisector

(iii) An isosceles triangle has only one line of symmetry. The line of symmetry is the bisectors of the angle contained by equal sides.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 30 Revision Exercise Symmetry image - 4
(iv) An equilateral triangle has three lines of symmetry. The bisectors of the angles are the lines of symmetry.

(v) A kite shaped figure has only one line of symmetry.

(vi) A figure of the shape of an arrow-head has only one line of symmetry.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 30 Revision Exercise Symmetry image - 7
(vii) The letter ‘A’ has only one line of symmetry.

(viii) The letter ‘H’ has two lines of symmetry.

(ix) A circle has an infinite number of lines of symmetry.
Every line, which passes through the centre of the circle, is a line of symmetry.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 30 Revision Exercise Symmetry image - 10

Revision Exercise Symmetry Exercise 30 – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
State, whether true or false :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 30 Revision Exercise Symmetry image - 11
Solution:
(If both the semi-circles of B are equal)
(i) True
(ii) True
(iii) False
(iv) False
(v) False
(vi) True
(vii) True
(viii) False.

Question 2.
Construct a triangle ABC, in which AB = AC = 5 cm and BC 6 cm. Draw all its lines of Symmetry.
Solution:
Steps of Construction :
(i) Draw a line BC = 6 cm.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 30 Revision Exercise Symmetry image - 12
(ii) With B as centre and radius 5 cm draw
(iii) With C as centre and radius 5 cm draw on arc which cut the previous arc at a.
(iv) Join AB and AC.
(v) ∆ABC is the required triangle line of symmetry is shown in the figure.

Question 3.
Examine each of the following figures carefully, draw line(s) of symmetry in which ever figure possible :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 30 Revision Exercise Symmetry image - 13
Solution:

Question 4.
Construct a triangle XYZ, in which XY = YZ = ZX = 4.5 cm. Draw all its lines of symmetry.
Solution:
Steps of Construction :
(i) Draw line XY = 4.5 cm
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 30 Revision Exercise Symmetry image - 16
(ii) With X as a centre and radius 4.5 cm draw an arc at Z.
(iii) With Y as a centre and radius 4.5 cm draw an arc which cuts the previous arc at Z.
(iv) Join XZ and YZ.
∆XYZ is the required triangle and lines of symmetries are shown in the figure.

Question 5.
Construct a triangle ABC, in which AB = BC = 4 cm and ∠ABC = 60°. Draw all its lines of symmetry.
Solution:
Steps of Construction :
(i) Draw a line AB = 4 cm.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 30 Revision Exercise Symmetry image - 17
(ii) At B draw an angle of 60° with the help of compass.
(iii) With B as centre and radius upon cut BC = 4 cm.
(iv) Join AC.
∆ABC is the required triangle and line of symmetry is shown in centre.

Question 6.
Draw the line(s) of symmetry for each figure drawn below :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 30 Revision Exercise Symmetry image - 18
Solution:

Question 7.
In each of the following case, construct a point that is symmetric to the given point P with respect to the given line AB.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 30 Revision Exercise Symmetry image - 20
Solution:

(i) Draw PQ perpendicular to AB from PO produced cut OQ = OP.
Point Q is symmetric to the given point P with respect to the given line AB.
(ii) Draw PO perpendicular to AB from P and produce PO to Q such that OQ = PO.
Point Q is symmetric to the given point P with respect to the given line AB.
(iii) Draw PQ perpendicular to AB from PO produced cut OQ = OP.
Point Q is symmetric to the given point P with respect to the given line AB.

Question 8.
Mark two points A and B 5.5 cm apart. Draw a line PQ so that A and B are symmetric with respect to the line PQ. Give a special name to line PQ.
Solution:
Steps of Construction :

Take points A and B 5.5 cm apart draw perpendicular bisect of the line segment AB. The perpendicular bisector PQ is the required line of symmetry with respect to A and B.
PQ is the perpendicular bisector of AB.

Question 9.
For each letter of the English alphabet, draw the maximum possible number of lines of symmetry.
Solution:

Question 10.
Draw all the possible lines of symmetry for each figure given below :

Solution:
The line or lines of symmetry of tile given figure are drawn as given below :
(i) A parallelogram has no line of symmetry.
(ii) A rectangle has two lines of symmetry.
(iii) A square has form lines of symmetry.

(iv) A semicircle has one line of symmetry.
(v) A quadrant has one line of symmetry.

Question 11.
For each shaded portion given below, draw all the possible lines of symmetry :
Solution:
The shaded portion of each figure given has the line or lines of symmetry as given below

(i) A minor segment has one line of symmetry.
(ii) A major segment has one line of symmetry.
(iii) quadrant has one line of symmetry.
(iv) The given figure has one line of symmetry.

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 32 Perimeter and Area of Plane Figures

August 11, 2022May 29, 2018 by Prasanna

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 32 Perimeter and Area of Plane Figures

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 32 Perimeter and Area of Plane Figures

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

Selina Class 6 Maths ICSE Solutions Physics Chemistry Biology Geography History & Civics

IMPORTANT POINTS

1. Perimeter: It is the length of the boundry of the given figure.
(i) Perimeter of a triangle = Sum of its three sides.
(ii) Perimeter of rectangle = 2 (length + breadth)
(iii) Perimeter of square = 4 x side.
2. Area: Area is the measure of surface of the plane covered by a closed plane figure. In other words, we can say that area of a closed plane figure is the measure of its interior region.
(i) Area of rectangle = length x breadth
(ii) Area of square = (side)².
3. Units of measurement of perimeter and area :
(i) Perimeter is measured in centimetre (cm) metre (m) or millimetre (mm).
(ii) Area is measured in square mm, square cm or square metre.

Perimeter and Area of Plane Figures Exercise 32A – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
What do you understand by a plane closed figure?
Solution:
Any geometrical plane figure bounded by lines (straight or curved) in a plane is called a plane closed figure.
Each of the following figures is a plane closed figure.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 32 Perimeter and Area of Plane Figures image - 1

Question 2.
The interior of a figure is called region of the figure. Is this statement true ?
Solution:
Yes. The interior of the figure alongwith its boundary is called region of the figure

Question 3.
Find the perimeter of each of the following closed figures :
Solution:
(i) Required perimeter
= AB + AC + BE + EF + FH + HG + HD
= 15 + 5 + 25 + 10 + 5 + 15 + 25 = 110 cm
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 32 Perimeter and Area of Plane Figures image - 2
(ii) Required perimeter
= AB + AC + CD + DG + BF + EF + EH + GH
= 20 + 4 + 8 + 20 + 4 + 8 + 20 + 4 = 88 cm

Question 4.
Find the perimeter of a rectangle whose:
(i) length = 40 cm and breadth = 35 cm
(ii) length = 10 m and breadth = 8 m
(iii) length = 8 m and breadth = 80 cm
(iv) length = 3.6 m and breadth = 2.4 m
Solution:
(i) length = 40 cm and breadth = 35 cm
∴Perimeter = 2 (length + breadth)
= 2 (40 cm + 35 cm)
= 2 x 75 cm
= 150 cm = \(\frac { 150 }{ 100 }\)
= 1.5 m
(ii) length = 10 m and breadth = 8 m
∴Perimeter = 2 (length + breadth)
= 2 (10 m + 8 m)
= 2 x 18 m = 54 m
(iii) length = 8 m and
breadth = 80 cm
Length = 8 m
Breadth = 80 cm= \(\frac { 80 }{ 100 }\) m = 0.8 m
∴ Perimeter = 2 (length + breadth)
= 2 (8 m + 0.8m)
= 2 x 8.8 m = 17.6 m
(iv) length = 3.6 m and breadth = 2.4 m
∴ Perimeter = 2 (length + breadth)
= 2 (3.6 m + 2.4 m)
= 2 x 6 m = 12 m

Question 5.
If P denotes perimeter of a rectangle, l denotes its length and b denotes its breadth, find :
(i) l, if P = 38cm and b = 7cm
(ii) b, if P = 3.2m and l = 100 cm
(iii) P, if l = 2 m and b = 75cm
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 32 Perimeter and Area of Plane Figures image - 4

Question 6.
Find the perimeter of a square whose each side is 1.6 m.
Solution:
∵ Side of the square = 1.6 m
∴ its perimeter = 4 x side
= 4 x 1.6 m
= 6.4 m

Question 7.
Find the side of the square whose pe-rimeter is 5 m.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 32 Perimeter and Area of Plane Figures image - 6

Question 8.
A square field has each side 70 m whereas a rectangular field has length = 50 m and breadth = 40 m. Which of the two fields has greater perimeter and by how much?
Solution:
Perimeter of the square field = 4 x side = 4 x 70m = 280m
Perimeter of rectangular field = 2 (length + breadth)
= 2 (50 m + 40 m)
= 2 x 90 m
= 180 m
∴Square field has greater perimeter by 280 m – 180 m = 100 m

Question 9.
A rectangular field has length = 160m and breadth = 120 m. Find :
(i) the perimeter of the field.
(ii) the length of fence required to enclose the field.
(iii) the cost of fencing the field at the rate of ? 80 per metre.
Solution:
Given = length = 160 m, breadth = 120m
(i) The Perimeter of the field = 2 (l + b)
= 2 (160 m + 120 m)
= 2 x 280
= 560 m
(ii) The length of fence required to enclose the field = The perimeter of the rectan-gular field
= 560 m
(iii) The cost of fencing the field = Length of fence x Rate of fence
= 560 m x ₹80 per metre
= ₹44, 800

Question 10.
Each side of a square plot of land is 55 m. Find the cost of fencing the plot at the rate of ₹32 per metre.
Solution:
∵Perimeter of square field = 4 x its side = 4 x 55 m
∴Length of required fencing = 220 m Now, the cost of fencing = its length x its rate
= 220 m x ₹32 per metre?
= ₹7040

Question 11.
Each side of a square field is 70 cm. How much distance will a boy walk in order to make ?
(i) one complete round of this field ?
(ii) 8 complete rounds of this field ?
Solution:
(i) Distance covered by the boy to make one complete round of the field.
Perimeter of the field : 4 x its side = 4 x 70 = 280 m
(ii) Distance covered by the boy to make 8 complete rounds of this field.
= 280 m x 8 m = 2240 m

Question 12.
A school playground is rectangular in shape with length = 120 m and breadth = 90 m. Some school boys run along the boundary of the play-ground and make 15 complete rounds in 45 minutes. How much distance they run during this period.
Solution:
Length of the rectangular playground = 120 mBreadth of the rectangular playground = 90 m
∴ Perimeter of the rectangular ground = 2(l + b)
= 2(120 + 90) m = 420 m
Thus, in one complete round, boys covers a distance of = 420 m
∴Distance covered in 15 complete rounds = 420 m x 15 = 6300 m

Question 13.
Mohit makes 8 full rounds of a rect-angular field with length = 120 m and breadth = 75 m.
John makes 10 full rounds of a square field with each side 100 in. Find who covers larger distance and by how much?
Solution:
Mohit
Length of the rectangular field = 120
Breadth of the rectangular field = 75 m
∴ Distance covered in one round (perim-eter) = 2(1 + b)
= 2(120 + 75) = 390 m Hence, distance covered in 8 rounds = 390 x 8 m = 3120 m
John
Side of the field = 100 m
∴Distance covered in one round = 4 x a = 4 x 100 = 400 m
Hence, Distance covered in 10 rounds = 400 x 10 m = 400 m
John a covers greater distance then Mohit by = (4000-3120) m = 880 m

Question 14.
The length of a rectangle is twice of its breadth. If its perimeter is 60 cm, find its length.
Solution:
Let the breadth of the field = x cm
∴ its length = 2x
and, its perimeter = 2 x (length + breadth)
= 2 x (2x + x)
= 2(3x)
= 6x cm
Perimeter = 60 cm
⇒ 60 cm = 6x cm
⇒ x = \(\frac { 60 }{ 6 }\) = 10 cm
∴Breadth = x = 10 cm
Length = 2x = 2 x 10 = 20 cm

Question 15.
Find the perimeter of :
(i) an equilateral triangle of side 9.8 cm.
(ii) an isosceles triangle with each equal side = 13 cm and the third side = 10 cm.
(iii) a regular pentagon of side 8.2 cm.
(iv) a regular hexagon of side 6.5 cm.
Solution:
(i) The perimeter of equilateral triangle = 3 x side
= 3 x 9.8 cm
= 29.4 cm
(ii) Required perimeter = 13 cm + 13 cm + 10 cm
= 36 cm
(iii) Perimeter of given pentagon = 5 x side = 5 x 8.2 cm
= 41 cm
(iv) Perimeter of given hexagon = 6 x side = 6 x 6.5 cm
= 39 cm

Question 16.
An equilateral triangle and d square has equal perimeter. If side of the triangle is 9.6 cm ; what is the length of the side of the square ?
Solution:
Perimeter of equilateral triangle = Perimeter of square Side of triangle = 9.6 cm
∴Perimeter of triangle = 3 x side
= 3 x 9.6 cm = 28.8 cm
> Perimeter of the square = 28.8 cm
4 x the side of square = 28.8 cm
⇒ The side of the square = \(\frac { 28.8 }{ 4 }\) cm
= 7.2 cm Ans.

Question 17.
A rectangle with length = 18 cm and breadth = 12 cm has same perimeter as that of a regular pentagon. Find the side of the pentagon.
Solution:
Length of rectangle = 18 cm
Breadth of rectangle = 12 cm
∴ Perimeter of rectangle = 2 x (l + b)
= 2 x (18+12)
= 2 x 30 = 60 cm
∵Perimeter, of rectangle = Perimeter of pentagon
60 cm = 5 x side
side = \(\frac { 60 }{ 5 }\) cm = 12 cm
∴Side of the pentagon = 12 cm Ans.

Question 18.
A regular pentagon of each side 12 cm has same perimeter as that of a regular hexagon. Find the length of each side of the hexagon.
Solution:
Perimeter of regular pentagon = 5 x length of the side
= 5 x 12 cm = 60 cm
Clearly, perimeter of the given pentagon = 60 cm
⇒ 6 x side of hexagon = 60 cm 60
⇒ side of hexagon = \(\frac { 60 }{ 6 }\)cm = 10 cm

Question 19.
Each side of a square is 45 cm and a rectangle has length 50 cm. If the perimeters of both (square and rectangle) are same, find the breadth of the rectangle.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 32 Perimeter and Area of Plane Figures image - 7

Question 20.
A wire is bent in the form of an equilateral triangle of each side 20 cm. If the same wire is bent in the form of a square, find the side of the square.
Solution:
∵Each side of the given equilateral triangle = 20 cm
∴Perimeter of the triangle = 3 x side = 3 x 20 cm = 60 cm ,
∴ Perimeter of the square = Perimeter of equilateral triangle
⇒ 4 x side of square = 60 cm
⇒ The side of the square = \(\frac { 60 }{ 4 }\)
=15 cm

Perimeter and Area of Plane Figures Exercise 32B – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Find the area of a rectangle whose :
(i) length = 15 cm breadth = 6.4 cm
(ii) Length = 8.5 m breadth = 5 m
(iii) Length = 3.6 m breadth = 90 cm
(iv) Length = 24 cm breadth =180 mm
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 32 Perimeter and Area of Plane Figures image - 8

Question 2.
Find the area of a square, whose each side is :
(i) 7.2 cm
(ii) 4.5 m
(iii) 4.1 cm
Solution:
(i) 7.2 cm
Area of the square = (side)² = (7.2 cm)² = 7.2 cm x 7.2 cm = 51.84 cm²
(ii) 4.5 m
Area of the square = (side)² = (4.5 m)² = 4.5 m x 4.5 m = 20.25 m²
(iii) 4.1 cm
Area of the square = (side)² = (4.1 cm)² = 4.1 cm x 4.1 cm = 16.81 cm²

Question 3.
If A denotes area of a rectangle, l represents its length and b represents its breadth, find :
(i) l, if A = 48 cm² and b = 6 cm
(ii) b, if A = 88 m² and l = 8m
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 32 Perimeter and Area of Plane Figures image - 9

Question 4.
Each side of a square is 3.6 cm; find its
(i) perimeter
(ii) area.
Solution:
(i) Perimeter = 4 x side
= 4 x 3.6 cm = 14.4 cm
(ii) Area = (side)²
= (3.6 cm)²
= 12.96 cm²

Question 5.
The perimeter of a square is 60 m, find :
(i) its each side its area
(ii) its new area obtained on increasing
(iii) each of its sides by 2 m.
Solution:
Perimeter of a square = 60 m
(i) Perimeter of a square = 4 x side
60 m = 4 x side
\(\frac { 60 }{ 4 }\) = side 4
∴side = 15 m
(ii) Area of square = (side)² = (15 m)²
= 15 m x 15 m
= 225 m²
(iii) Increased each side = 2 m
Side of square = 15 m
New length of side = (2m + 15m)
= 17m
∴New Area of square = (17m)² = 17m x 17m = 289 m²

Question 6.
Each side of a square is 7 m. If its each side be increased by 3 m, what will be the increase in its area.
Solution:
Each side of square = 7 m
∴Area of square = (side)²= (7 m)²
= 7m x 7m =49m²
∵ Side increased by 3 m
∴Total length of side will be = 3 m + 7 m = 10m
∴Area of square = (10 m)²= 10m x 10 m = 100 m²
∴Increase in area = 100 m² – 49 m² = 51 m²

Question 7.
The perimeter of a square field is numerically equal to its area. Find each side of the square.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 32 Perimeter and Area of Plane Figures image - 10

Question 8.
A rectangular piece of paper has area = 24 cm² and length = 5 cm. Find its perimeter.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 32 Perimeter and Area of Plane Figures image - 11

Question 9.
Find the perimeter of a rectangle whose area = 2600 m² and breadth = 50 m.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 32 Perimeter and Area of Plane Figures image - 12

Question 10.
What will happen to the area of a rectangle, if its length and breadth both are trebled?
Solution:
Let the original length of the rectangle = l and its original breadth = b
∴ its original area = length x breadth i.e A = l – b i. e.
Since,
Increased length -=3l
and, increased breadth = 3b
∴ New area = 3l x 3b = 9 x l x b [∵A = l x b]
⇒ Area of the new rectangle = 9 times than area of original rectangle

Question 11.
Length of a rectangle is 30 m and its breadth is 20 m. Find the increase in its area if its length is increased by 10 m and its breadth is doubled.
Solution:
Length of a rectangle (l) = 30 m,
Breadth of the rectangle (b) = 20 m
Area of rectangle = l x b
= 30 x 20 = 600 m2
Since, the length its increased by 10 m and breadth is doubled
∴New length (l) = (30 + 10) m = 40 m
and new breadth = (20 x 2) m = 40 m
∴New area = l x b = 40 x 40 m2 = 1600 m2
Hence, the increase in the area = (1600 – 600) m2
= 1000 m2

Question 12.
The side of a square field is 16 m. What will be increase in its area, if:
(i) each of its sides is increased by 4 m
(ii) each of its sides is doubled.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 32 Perimeter and Area of Plane Figures image - 13

Question 13.
Each rectangular tile is 40 cm long and 30 cm wide. How many tiles will be required to cover the floor of a room with length = 4.8 m and breadth = 2.4 m.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 32 Perimeter and Area of Plane Figures image - 15

Question 14.
Each side of a square tile is 60 cm. How many tiles will be required to cover the floor of a hall with length = 50 m and breadth = 36 m.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 32 Perimeter and Area of Plane Figures image - 16

Question 15.
The perimeter of a square plot = 360 m. Find :
(i) its area.
(ii) cost of fencing its boundary at the rate of ₹ 40 per metre.
(iii) cost of levelling the plot at ₹60 per square metre.
Solution:
Given, perimeter of square plot = 360 m
∵ Perimeter of the square = 4 x its side
∴4 x side of square = 360 m
⇒ side of the square = \(\frac { 360m }{ 4 }\) = 90 m
(i) The area of the square field = (side)²
= (90 m)²
= 90 m x 90 m
= 8100 m²
Cost of fencing at ₹ 40 per metre
= 8100 m2 x ₹ 40 per metre
= ₹ 324000
Cost of levelling at₹ 60 per m²
= 8100 m² x ₹60 per m²
= ₹ 486000

Question 16.
The perimeter of a rectangular field is 500 m and its length = 150 m. Find:
(i) its breadth,
(ii) its area.
(iii) cost of ploughing the field at the rate of ₹1.20 per square metre.
Solution:
(i) Perimeter of a rectangle = 2 x (length + breadth)
⇒500 m = 2x(i50m + breadth)
⇒250 m – 150 m = breadth
∴breadth = 100 m
(ii) Area of rectangular field = length x breadth
= 150 m x 100 m = 15000 m²
(iii) Cost of ploughing the field at the rate of
= ₹1.20 per square m²= area of the field x rate of ploughing = 15000 m² x ₹1.20 per square metre = ₹15000 x 1.20 = ₹18000

Question 17.
The cost of flooring a hall of ₹64 per square metre is ₹2,048. If the breadth of the hall is 5m, find :
(i) its length.
(ii) its perimeter.
(iii) cost of fixing a border of very small width along its boundary at the rate of ₹60 per square metre.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 32 Perimeter and Area of Plane Figures image - 17

Question 18.
The length of a rectangle is three times its breadth. If the area of the rectangle is 1875 sq. cm, find its perimeter.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 32 Perimeter and Area of Plane Figures image - 20

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 31 Recognition of Solids

August 11, 2022May 29, 2018 by Prasanna

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 31 Recognition of Solids

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 31 Recognition of Solids

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

Selina Class 6 Maths ICSE Solutions Physics Chemistry Biology Geography History & Civics

Recognition of Solids Exercise 31 – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Identify the nets which can be used to form cubes :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 31 Recognition of Solids image - 1
Solution:
Nets for a cube are (ii), (iii) and (iv).

Question 2.
Draw at least three different nets for making cube.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 31 Recognition of Solids image - 2

Question 3.
The dimensions of a cuboid are 6 cm, 4 cm and 3 cm. Draw two different types of oblique sketches for this cuboid.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 31 Recognition of Solids image - 3

Question 4.
Two cubes, each with 3 cm edge, are placed side by side to form a cuboid. For this cuboid, draw :
(i) an oblique sketch
(ii) an isometric sketch.

Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 31 Recognition of Solids image - 4

Question 5.
The figure, given below, shows shadows of some 3D objects, when seen under the lamp of an overhead projector :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 31 Recognition of Solids image - 6
In each case, name the object.
Solution:

Question 6.
Look at the solids, drawn below, and fill the given chart.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 31 Recognition of Solids image - 8

Solution:

P. Q. Using Euler’s formula, find the values of a, b, c and d.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 31 Recognition of Solids image - 11

Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 31 Recognition of Solids image - 12
(i) a + 6 — 12 = 2 ⇒a = 2- 6+12 = 14-6 = 8
(ii) b + 5- 9 = 2 ⇒b-2 + 9-5 = 6
(iii) 20 + 12 – c = 2 ⇒ 32 – c = 2 ⇒ c = 32 – 2 ⇒ c = 30
(iv) 6 + d – 12 = 2⇒ d-6 = 2 ⇒ d = 2 + 6 = 8

P.Q. Using an isometric dot paper, draw :
(i) a cube with each edge 3 cm.
(ii) a cuboid measuring 5 cm x 4 cm x 3 cm.

Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 31 Recognition of Solids image - 13

Question 7.
Dice are cubes with dot or dots on each face. Opposite faces of a die always have a total of seven on them.
Below are given two nets to make dice (cube), the numbers inserted in each square indicate the number of dots in it.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 31 Recognition of Solids image - 15
Insert suitable numbers in each blank so that numbers in opposite faces of the die have a total of seven dots.
Solution:

Question 8.
The following figure represents mets of some solids. Name the solids.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 31 Recognition of Solids image - 17
Solution:
The given nets are of the solid as given below :
(i) Tetrahedron
(ii) Triangular prism
(iii) Cube
(iv) Cuboid

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 33 Data Handling

August 11, 2022May 26, 2018 by Prasanna

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 33 Data Handling (Including Pictograph and Bar Graph)

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 33 Data Handling (Including Pictograph and Bar Graph)

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

Selina Class 6 Maths ICSE Solutions Physics Chemistry Biology Geography History & Civics

IMPORTANT POINTS

The word statistics is used for two different meanings.
In the singular sense, it is used as a science or a subject which deals with the collection, classification, tabulation, representation and interpretation of the data.
In the plural sense, it is sometimes used for the numerical facts collected in the form of numbers.
If we have collected information about the heights of class 6 children from ten different schools of Delhi, then this information in the form of numbers is called statistics.
1. Data : Each number, collected for giving a required information, is called the data.

2. Bar Graph (Column Graph) : Bar graph is the simplest form of presenting a data. It consists of bars (usually vertical), all of same widths. The heights of these bars are drawn according to the number they represent.

3. Pie Graph : When the given data is represented by the sectors of a circle, the resulting diagram (graph) obtained is called a pie-graph or a pie-chart.

Data Handling Exercise 33A – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Marks scored by 30 students of class VI are as given below :
38, 46, 33, 45, 63, 53, 40, 85, 52, 75, 60, 73, 62, 22, 69, 43, 45, 33, 47, 41, 29, 43, 37, 49, 83, 44, 55, 22, 35 and 45. State :
(i) the highest marks scored.
(ii) the lowest marks scored.
(iii) the range of marks.
Solution:
(i) Highest marks scored = 85 .
(ii) Lowest marks scored = 22
(iii) Range of marks = 85 – 22 = 63

Question 2.
For the following raw data, form a discrete frequency distribution :
30,32,32, 28,34,34,32,30,30,32,32,34,30,32,32. 28,32,30, 28,30,32,32,30,28 and 30.
Solution:
The required frequency table will be as shown below :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 33 Data Handling image - 1

Question 3.
Define :
(i) data
(ii) frequency of an observation.
Solution:
(i) Data : The word data means information in the form of numerical figures.
(ii) Frequency of an observation : The number of times a particular observation occurs is called its frequency.

Question 4.
Rearranage the following raw data in descending order :
5.3, 5.2, 5.1, 5.7, 5.6, 6.0, 5.5, 5.9, 5.8, 6.1, 5.5, 5.8, 5.7, 5.9 and 5.4. Then write the :
(i) highest value
(ii) lowest value
(iii) range of values
Solution:
Writing these numbers in descending order we get:
6.1, 6.0, 5.9, 5.9, 5.8, 5.8, 5.7, 5.7, 5.6, 5.5, 5.5, 5.4, 5.3, 5.2, 5.1
(i) Highest value = 6.1
(ii) Lowest value = 5.1
(iii) Range of values = Highest value – lowest value = 6.1 -5.1 = 1.0

Question 5.
Represent the following data in the form of a frequency distribution :
52, 56, 72, 68, 52, 68, 52, 68, 52, 60, 56, 72, 56, 60, 64, 56, 48, 48, 64 and 64.
Solution:
The required frequency table wilf be as shown below :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 33 Data Handling image - 2

Question 6.
In a study of number of accidents per day, the observations for 30 days were obtained as follows :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 33 Data Handling image - 3
Solution:
The required frequency table will be as shown below :

Question 7.
The following data represents the weekly wages (in ₹) of 15 workers in a factory : 900, 850, 800, 850, 800, 750, 950, 900, 950, 800, 750, 900, 750, 800 and 850.
Prepare a frequency distribution table. Now find,
(i) how many workers are getting less than ₹850 per week?
(ii) how many workers are getting more than ₹800 per week?
Solution:
The required frequency table will be as shown below :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 33 Data Handling image - 5
(i) Workers getting less than ₹850 per week
No. of workers getting ₹750 = 3 workers
No. of workers getting ₹800 = 4 works
∴ Workers getting less than ₹ 850 = 4 + 3 = 7 workers
(ii) Workers are getting more than ₹800 per week
No. of workers getting ₹850 = 3
No. of workers getting ₹900 = 3
No. of workers getting ₹950 = 2
∴Workers getting more than ₹800 = 3 + 3 + 2 = 8 workers

Question 8.
Using the data, given below, construct a frequency distribution table : 9, 17, 12, 20, 9, 18, 25, 17, 19, 9, 12, 9, 12, 18, 17, 19, 20, 25, 9 and 12. Now answer the following :
(i) How many numbers are less than 19?
(ii) How many numbers are more than 20?
(iii) Which of the numbers, given above, is occuring most frequently?
Solution:
The required frequency table will be as shown below :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 33 Data Handling image - 6
(i) There are 14 numbers are less than 19.
(ii) There are 2 numbers more than 20.
(iii) 9 is occuring most frequently i.e. 5 times.

Question 9.
Using the following data, construct a frequency distribution table : 46, 44, 42, 54, 52, 60, 50, 58, 56, 62, 50, 56, 54, 58 and 48.
Now answer the following :
(i) What is the range of the numbers?
(ii) How many numbers are greater than 50?
(iii) How many numbers are between 40 and 50?
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 33 Data Handling image - 7
(i) Range of numbers = Highest number – Lowest number = 62 – 42 = 20
(ii) 9 numbers are greater than 50
(iii) 6 numbers are between 40 and 50 Ans.

Data Handling Exercise 33B – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
The sale of vehicles, in a particular city, during the first six months of the year 2016 is shown below :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 33 Data Handling image - 8
vehicles sold
Draw a pictograph to represent the above data.
Solution:

Question 2.
Selina Concise Mathematicsclass 6 ICSE Solutions -Data Handling (Including Pictograph and Bar Graph) - 2b
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 33 Data Handling image - 10
Solution:
(i) Cars sold by dealer A = 6 x 50 = 300
Cars sold by dealer D = 4 x 5 = 200 ,
∴ A sold more cars than dealer D by = 300 – 200 = 100
∴A has sold 100 more cars than dealer D.
(ii) No. of cars = 23
Scale = 50 cars
∴Total no. of cars = 23 x 50 = 1150 cars Ans.

Question 3.
The following pictograph shows the number of watches manufactured by a factory, in a particular weeks.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 33 Data Handling image - 11

Find
(i) on which day were the least number of w atches manufactured ?
(ii) total number of watches manufatured in the whole week ?
Solution:
(i) On Friday least no. of watches manufactured by = 100 x 5 = 500 watches
(ii) Total no. of watches manufactured in the whole week = 100 x 42.5 = 4250 watches

Question 4.
The number or animals in five villages are as follows :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 33 Data Handling image - 13
Prepare a pitctograph of these animals using one symbol to represent 20 animals.
Solution:

Question 5.
The following pictograph shows different subject books which are kept in a school library.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 33 Data Handling image - 15
Taking symbol of one book = 50 books, find :
(i) how many History books are there in the library ?
(ii) how many Science books are there in the library ?
(iii) which books are maximum in number ?

Solution:
(i) There are 50 x 4 = 200 History books in the library.
(ii) There are 50 x 5.5 = 275 Science books in the library.
(iii) English books are maximum in number = 500 x 9 = 450 books.

Data Handling Exercise 33C – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
The following table gives the number of students in class VI in a school during academic years 2011-2012 to 2015-2016.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 33 Data Handling image - 16
Represent the above data by a bar graph.
Solution:

Question 2.
The attendence of a particular class for the six days of a week are as given below :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 33 Data Handling image - 18
Draw a suitable graph.
Solution:

Question 3.
The total number of students present in class VI B, for the six days in a week were as given below. Draw a suitable bar graph.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 33 Data Handling image - 20
Solution:

Question 4.
The following table shows the population of a particular city at different years :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 33 Data Handling image - 22
Represent the above information with the help of a suitable bar graph.
Solution:

Question 5.
In a survey of 300 families of a colony, the number of children in each family was recorded and the data has been represented by the bar graph, given below :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 33 Data Handling image - 24
Read the graph carefully and answer the following questions :
(i) How many families have 2 children each ?
(ii) How many families have no child ?
(iii) What percentage of families have 4 children ?
Solution:
(i) 60 families have 2 children each.
(ii) Zero
(iii) The percentage of families having 4 children = \(\frac { 60 }{ 300 }\) x 100 = 20%

Question 6.
Use the data, given in the following table, to draw’ a bar graph
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 33 Data Handling image - 25
Out of A, B, C, D, E and F
(i) Which has the maximum value.
(ii) Which is greater A + D or B + E.
Solution:
(i) D has the maximum value of 350
(ii) A + D = 250 + 350 = 600
B + E = 300 + 275 = 575
Hence A + D is greater.

Question 7.
The bar graph drawn below shows the number of tickets sold during a fair by 6 students A, B, C, D, E and F.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 33 Data Handling image - 26
Using the Bar graph, answer the following questions :
(i) Who sold the least number of tickets?
(ii) Who sold the maximum number of tickets ?
(iii) How many tickets were sold by A, B and C taken together ?
(iv) How many tickets were sold by D, E and F taken together ?
(v) What is the average number of tickets sold per student ?
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 33 Data Handling image - 27

Question 8.
The following bar graph shows the number of children, in various classes, in a school in Delhi.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 33 Data Handling image - 29
Using the given bar graph, find :
(i) the number of children in each class.
(ii) the total number of children from Class 6 to Class 8.
(iii) how many more children there are in Class 5 compared to Class 6 ?
(iv) the total number of children from Class 1 to Class 8.
(v) the average number of children in a class.

Solution:
(i) In, Class 1 = 100, Class 2 = 90, Class 3 = 100, Class 4 = 80,
Class 5 = 120, Class 6 = 90, Class 7 = 70, Class 8 = 50
(ii) Class 6 = 90, Class 7 = 70, Class 8 = 50, Total number = 210
(iii) Number of student in class 5 = 120, Number of student in class 6 = 90
More children is class 5 = (120 – 90) = 30
(iv) Total number of children class 1 to 8 = 100 + 90+ 100+ 80 + 120 +90 + 70 + 50 = 700

Question 9.
The column graph, given above , shows the number of patients, examined by Dr. V.K. Bansal, on different days of a particular week.
Use the graph to answer the following:
(i) On which day were the maximum number of patients examined ?
(ii) On which day were the least number of patients examined ?
(iii) On which days were equal number or patients examined ?
(iv) What is the total number of patients examined in the week ?
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 33 Data Handling image - 31
(i) Tuesday were the maximum number of patients examined.
(ii) Friday were the least number of patients examined.
(iii) Sunday and Thursday were equal number of patient examined.
(iv) Total number of patients examined in the week .
= 50 + 40 + 70 + 60 + 50 + 30 = 300

Question 10.
A student spends his pocket money on various items, as given below :
Books : Rs. 380, Postage : Rs. 30, Cosmetics : Rs. 240, Stationary : Rs. 220 and Entertainment : Rs. 120.
Draw a bar graph to represent his expenses.
Solution:
Amount spent on
Books = Rs. 380
Postage = Rs. 30
Cosmetics = Rs. 240
Stationary = Rs. 220
Entertainment = Rs. 120
The bar graph of the above given data is below.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 33 Data Handling image - 32