## Selina Concise Mathematics Class 9 ICSE Solutions Distance Formula

APlusTopper.com provides step by step solutions for Selina Concise Mathematics Class 9 ICSE Solutions Chapter 28 Distance Formula. You can download the Selina Concise Mathematics ICSE Solutions for Class 9 with Free PDF download option. Selina Publishers Concise Mathematics for Class 9 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines

Selina ICSE Solutions for Class 9 Maths Chapter 28 Distance Formula

Exercise 28(A)

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## Selina Concise Mathematics Class 9 ICSE Solutions Statistics

APlusTopper.com provides step by step solutions for Selina Concise Mathematics Class 9 ICSE Solutions Chapter 18 Statistics. You can download the Selina Concise Mathematics ICSE Solutions for Class 9 with Free PDF download option. Selina Publishers Concise Mathematics for Class 9 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

Selina ICSE Solutions for Class 9 Maths Chapter 18 Statistics

Exercise 18(A)

Solution 1:
(a) Discrete variable.
(b) Continuous variable.
(c) Discrete variable.
(d) Continuous variable.
(e) Discrete variable.

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In this frequency distribution, the marks 30 are in the class of interval 30 – 40 and not in 20 – 30. Similarly, marks 40 are in the class of interval 40 – 50 and not in 30 – 40.

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Exercise 18(B)

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## Selina Concise Mathematics Class 9 ICSE Solutions Solution of Right Triangles [Simple 2-D Problems Involving One Right-angled Triangle]

APlusTopper.com provides step by step solutions for Selina Concise Mathematics Class 9 ICSE Solutions Chapter 24 Solution of Right Triangles [Simple 2-D Problems Involving One Right-angled Triangle]. You can download the Selina Concise Mathematics ICSE Solutions for Class 9 with Free PDF download option. Selina Publishers Concise Mathematics for Class 9 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

Selina ICSE Solutions for Class 9 Maths Chapter 24 Solution of Right Triangles [Simple 2-D Problems Involving One Right-angled Triangle]

Exercise 24

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Thus AB = AC + CD + BD = 54.64 cm.

Solution 8:
First draw two perpendiculars to AB from the point D and C respectively. Since AB|| CD therefore PMCD will be a rectangle.
Consider the figure,

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Given that the ladder makes an angle of 30o with the ground and reaches upto a height of 15 m of the tower which is shown in the figure below:

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We also know that in rhombus diagonals bisect each other perpendicularly and diagonal bisect the angle at vertex.
Hence POR is a right angle triangle and

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## Selina Concise Mathematics Class 9 ICSE Solutions Rectilinear Figures [Quadrilaterals: Parallelogram, Rectangle, Rhombus, Square and Trapezium]

APlusTopper.com provides step by step solutions for Selina Concise Mathematics Class 9 ICSE Solutions Chapter 14 Rectilinear Figures [Quadrilaterals: Parallelogram, Rectangle, Rhombus, Square and Trapezium]. You can download the Selina Concise Mathematics ICSE Solutions for Class 9 with Free PDF download option. Selina Publishers Concise Mathematics for Class 9 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

Selina ICSE Solutions for Class 9 Maths Chapter 14 Rectilinear Figures [Quadrilaterals: Parallelogram, Rectangle, Rhombus, Square and Trapezium]

Exercise 14(A)

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In this linear equation n and k must be integer. Therefore to satisfy this equation the minimum value of k must be 6 to get n as integer.
Hence the number of sides are: 5 + 6 = 11.

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Exercise 14(B)

Solution 1:
(i)True.
This is true, because we know that a rectangle is a parallelogram. So, all the properties of a parallelogram are true for a rectangle. Since the diagonals of a parallelogram bisect each other, the same holds true for a rectangle.
(ii)False
This is not true for any random quadrilateral. Observe the quadrilateral shown below.

Clearly the diagonals of the given quadrilateral do not bisect each other. However, if the quadrilateral was a special quadrilateral like a parallelogram, this would hold true.
(iii)False
Consider a rectangle as shown below.

It is a parallelogram. However, the diagonals of a rectangle do not intersect at right angles, even though they bisect each other.
(iv)True
Since a rhombus is a parallelogram, and we know that the diagonals of a parallelogram bisect each other, hence the diagonals of a rhombus too, bisect other.
(v)False
This need not be true, since if the angles of the quadrilateral are not right angles, the quadrilateral would be a rhombus rather than a square.
(vi)True

A parallelogram is a quadrilateral with opposite sides parallel and equal.
Since opposite sides of a rhombus are parallel, and all the sides of the rhombus are equal, a rhombus is a parallelogram.
(vii)False
This is false, since a parallelogram in general does not have all its sides equal. Only opposite sides of a parallelogram are equal. However, a rhombus has all its sides equal. So, every parallelogram cannot be a rhombus, except those parallelograms that have all equal sides.
(viii)False
This is a property of a rhombus. The diagonals of a rhombus need not be equal.
(ix)True
A parallelogram is a quadrilateral with opposite sides parallel and equal.
A rhombus is a quadrilateral with opposite sides parallel, and all sides equal.
If in a parallelogram the adjacent sides are equal, it means all the sides of the parallelogram are equal, thus forming a rhombus.
(x)False

Observe the above figure. The diagonals of the quadrilateral shown above bisect each other at right angles, however the quadrilateral need not be a square, since the angles of the quadrilateral are clearly not right angles.

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We know that AQCP is a quadrilateral. So sum of all angles must be 360.
∴ x + y + 90 + 90 = 360
x + y = 180
Given x:y = 2:1
So substitute x = 2y
3y = 180
y = 60
x = 120
We know that angle C = angle A = x = 120
Angle D = Angle B = 180 – x = 180 – 120 = 60
Hence, angles of parallelogram are 120, 60, 120 and 60.

Exercise 14(C)

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## Selina Concise Mathematics Class 9 ICSE Solutions Area Theorems [Proof and Use]

APlusTopper.com provides step by step solutions for Selina Concise Mathematics Class 9 ICSE Solutions Chapter 16 Area Theorems [Proof and Use]. You can download the Selina Concise Mathematics ICSE Solutions for Class 9 with Free PDF download option. Selina Publishers Concise Mathematics for Class 9 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

Selina ICSE Solutions for Class 9 Maths Chapter 16 Area Theorems [Proof and Use]

Exercise 16(A)

Solution 1:

Solution 2:

Since from the figure, we get CD//FE therefore FC must parallel to DE. Therefore it is proved that the quadrilateral CDEF is a parallelogram.
Area of parallelogram on same base and between same parallel lines is always equal and area of parallelogram is equal to the area of rectangle on the same base and of the same altitude i.e, between same parallel lines.
So Area of CDEF= Area of ABDC + Area of ABEF
Hence Proved

Solution 3:

Solution 4:

Given ABCD is a parallelogram. P and Q are any points on the sides AB and BC respectively, join diagonals AC and BD.
proof:
since triangles with same base and between same set of parallel lines have equal areas
area (CPD)=area(BCD)…… (1)
again, diagonals of the parallelogram bisects area in two equal parts
area (BCD)=(1/2) area of parallelogram ABCD…… (2)
from (1) and (2)
area(CPD)=1/2 area(ABCD)…… (3)
similarly area (AQD)=area(ABD)=1/2 area(ABCD)…… (4)
from (3) and (4)
area(CPD)=area(AQD),
hence proved.
(ii)
We know that area of triangles on the same base and between same parallel lines are equal
So Area of AQD= Area of ACD= Area of PDC = Area of BDC = Area of ABC=Area of APD + Area of BPC
Hence Proved

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We know that area of triangles on the same base and between same parallel lines are equal.
With common base, BE and between AD and BE parallel lines, we have
Area of ΔABE = Area of ΔBDE
With common base BC and between BE and CF parallel lines, we have
Area of ΔBEC = Area of ΔBEF
Area of ΔABE + Area of ΔBEC = Area of ΔBEF + Area of ΔBDE
=> Area of AEC = Area of DBF
Hence Proved

Solution 17:
Given: ABCD is a parallelogram.
We know that
Area of ΔABC = Area of ΔACD
Consider ΔABX,
Area of ΔABX = Area of ΔABC + Area of ΔACX
We also know that area of triangles on the same base and between same parallel lines are equal.
Area of ΔACX = Area of ΔCXD
From above equations, we can conclude that
Area of ΔABX = Area of ΔABC + Area of ΔACX = Area of ΔACD+ Area of ΔCXD = Area of ACXD Quadrilateral
Hence Proved

Solution 18:
Join B and R and P and R.
We know that the area of the parallelogram is equal to twice the area of the triangle, if the triangle and the parallelogram are on the same base and between the parallels
Consider ABCD parallelogram:
Since the parallelogram ABCD and the triangle ABR lie on AB and between the parallels AB and DC, we have

Exercise 16(B)

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We have to join PD and BD.

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Ratio of area of triangles with same vertex and bases along the same line is equal to ratio of their respective bases. So, we have

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Exercise 16(C)

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## Selina Concise Mathematics Class 9 ICSE Solutions Trigonometrical Ratios [Sine, Consine, Tangent of an Angle and their Reciprocals]

APlusTopper.com provides step by step solutions for Selina Concise Mathematics Class 9 ICSE Solutions Chapter 22 Trigonometrical Ratios [Sine, Cosine, Tangent of an Angle and their Reciprocals]. You can download the Selina Concise Mathematics ICSE Solutions for Class 9 with Free PDF download option. Selina Publishers Concise Mathematics for Class 9 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines

Selina ICSE Solutions for Class 9 Maths Chapter 22 Trigonometrical Ratios [Sine, Consine, Tangent of an Angle and their Reciprocals]

Exercise 22(A)

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## Selina Concise Mathematics Class 9 ICSE Solutions Mid-point and Its Converse [ Including Intercept Theorem]

APlusTopper.com provides step by step solutions for Selina Concise Mathematics Class 9 ICSE Solutions Chapter 12 Mid-point and Its Converse [ Including Intercept Theorem]. You can download the Selina Concise Mathematics ICSE Solutions for Class 9 with Free PDF download option. Selina Publishers Concise Mathematics for Class 9 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

Selina ICSE Solutions for Class 9 Maths Chapter 12 Mid-point and Its Converse [ Including Intercept Theorem]

Exercise 12(A)

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Exercise 12(B)

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Given ABCD is parallelogram, so AD = BC, AB = CD.
Consider triangle APB, given EC is parallel to AP and E is midpoint of side AB. So by midpoint theorem, C has to be the midpoint of BP.
So BP = 2BC, but BC = AD as ABCD is a parallelogram.
Consider triangle APB, AB || OC as ABCD is a parallelogram. So by midpoint theorem, O has to be the midpoint of AP.
Hence Proved

Solution 13:
Consider trapezium ABCD.
Given E and F are midpoints on sides AD and BC, respectively.

Consider LHS,
AB + CD = AB + CJ + JI + ID = AB + 2HF + AB + 2EG
So AB + CD = 2(AB + HF + EG) = 2(EG + GH + HF) = 2EF
AB + CD = 2EF
Hence Proved

Solution 14:
Given Δ ABC
AD is the median. So D is the midpoint of side BC.
Given DE || AB. By the midpoint theorem, E has to be midpoint of AC.
So line joining the vertex and midpoint of the opposite side is always known as median. So BE is also median of Δ ABC.

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## Selina Concise Mathematics Class 9 ICSE Solutions Inequalities

APlusTopper.com provides step by step solutions for Selina Concise Mathematics Class 9 ICSE Solutions Chapter 11 Inequalities. You can download the Selina Concise Mathematics ICSE Solutions for Class 9 with Free PDF download option. Selina Publishers Concise Mathematics for Class 9 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

Selina ICSE Solutions for Class 9 Maths Chapter 11 Inequalities

Exercise 11

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## Selina Concise Mathematics Class 9 ICSE Solutions Logarithms

APlusTopper.com provides step by step solutions for Selina Concise Mathematics Class 9 ICSE Solutions Chapter 8 Logarithms. You can download the Selina Concise Mathematics ICSE Solutions for Class 9 with Free PDF download option. Selina Publishers Concise Mathematics for Class 9 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

Selina ICSE Solutions for Class 9 Maths Chapter 8 Logarithms

Exercise 8(A)

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## Selina Concise Mathematics Class 9 ICSE Solutions Circle

APlusTopper.com provides step by step solutions for Selina Concise Mathematics Class 9 ICSE Solutions Chapter 17 Circle. You can download the Selina Concise Mathematics ICSE Solutions for Class 9 with Free PDF download option. Selina Publishers Concise Mathematics for Class 9 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

Selina ICSE Solutions for Class 9 Maths Chapter 17 Circle

Exercise 17(A)

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Let O be the centre of the circle and AB and CD be the two parallel chords of length 30 cm and 16 cm respectively.
Drop OE and OF perpendicular on AB and CD from the centre O.

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Since the distance between the chords is greater than the radius of the circle (15 cm), so the chords will be on the opposite sides of the centre.

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Exercise 17(C)

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As given that AB is the side of a pentagon the angle subtended by each arm of the pentagon at

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## Selina Concise Mathematics Class 9 ICSE Solutions Area and Perimeter of Plane Figures

APlusTopper.com provides step by step solutions for Selina Concise Mathematics Class 9 ICSE Solutions Chapter 20 Area and Perimeter of Plane Figures. You can download the Selina Concise Mathematics ICSE Solutions for Class 9 with Free PDF download option. Selina Publishers Concise Mathematics for Class 9 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

Selina ICSE Solutions for Class 9 Maths Chapter 20 Area and Perimeter of Plane Figures

Exercise 20(A)

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Since the perimeter of the isosceles triangle is 36cm and base is 16cm.

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Exercise 20(B)

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We need to find the cost of carpeting of 80 cm = 0.8 m wide carpet, if the rate of carpeting is Rs. 25. Per metre.
Then

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In the given figure, we can observe that the non-parallel sides are equal and hence it is an isosceles trapezium.

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From the given data, we can calculate the area of the outer circle and then the area of inner circle and hence the width of the shaded portion.

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## Selina Concise Mathematics Class 9 ICSE Solutions Construction of Polygons (Using ruler and compass only)

APlusTopper.com provides step by step solutions for Selina Concise Mathematics Class 9 ICSE Solutions Chapter 15 Construction of Polygons (Using ruler and compass only). You can download the Selina Concise Mathematics ICSE Solutions for Class 9 with Free PDF download option. Selina Publishers Concise Mathematics for Class 9 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

Selina ICSE Solutions for Class 9 Maths Chapter 15 Construction of Polygons (Using ruler and compass only)

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Steps:

1. We construct the segment AC = 7cm.
2. With A as a centre and radius 5.4 cm , we draw an arc extending on both sides of AC.
3. With C as centre and same radius as in step 2, we draw an arc extending on both sides of AC to cut the first arc at B and D.
4. Join AB,BC,CD and DA.

ABCD is the required rhombus.

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Steps:

1. Draw a base line AQ.
2. From A take some random distance in compass and draw one are below and above the line. Now without changing the distance in compass draw one are below and above the line. These arcs intersect each other above and below the line.
3. Draw the line passing through these intersecting points, you will get a perpendicular to the line AQ.
4. Take distance of 4 cm in compass and mark an arc on the perpendicular above the line. Draw a line parallel to line AQ passing through through this arc.
5. From point A measure an angle of 60 degree and draw the line which intersect above drawn line at some point label it as D.
6. Using the procedure given in step 2 again draw a perpendicular to line AD.
7. Take distance of 3 cm in compass and mark an arc on the perpendicular above the line. Draw a line parallel to line AD passing through through this arc which intersect the line AQ at some point label it as B and to other line at point C.

ABCD is the required parallelogram.

Solution 38:
To draw the parallelogram follows the steps:

1. First draw a line AB of measure 6cm. Then draw an angle of measure 450 at point A such that ∠DAB = 450 and AD = 5cm.
2. Now draw a line CD parallel to the line AB of measure 6cm. Then join BC to construct the parallelogram as shown below:

Solution 39:

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