ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 15 Visualising Solid Shapes Ex 15.3

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 15 Visualising Solid Shapes Ex 15.3

Question 1.
Count the number of cubes in the following structures:
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 15 Visualising Solid Shapes Ex 15.3 Q1.1
Solution:
Number of cubes in the given structure havebeen given below:
(i) 5 × 3 × 2 = 30
(ii) 7 × 2 + 6 = 14 + 6 = 20
(iii) 5 × 5 + 17 + 4 = 25 + 17 + 4 = 46

Question 2.
What cross-section is made in the following
(i) vertical cut (ii) horizontal cut?
(a) A brick
(b) A round apple
(c) A die
(d) A circular pipe
(e) An ice cream cone
(f) A square pyramid
Solution:

Vertical cut Horizontal cut
(a) A brick Rectangle Rectangle
(b) A round apple Circle Circle
(c) A die Square Square
(d) A circular pipe Rectangle Circle
(e) An ice cream cone Triangle Circle
(f) A square pyramid Triangle Square

Question 3.
For each solid given below, the three views (1), (2) and (3) are given. Identify for each solid the corresponding top, front and side views:
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 15 Visualising Solid Shapes Ex 15.3 Q3.1
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 15 Visualising Solid Shapes Ex 15.3 Q3.2
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 15 Visualising Solid Shapes Ex 15.3 Q3.3
Solution:
(a) (1) Side (2) Top (3) Front
(b) (1) Front (Top) (2) Side (3) Top (Front)
(c) (1) Top (2) Side (3) Front
(d) (1) Front (Side) (2) Side (Front) (3) Top
(e) (1) Top (2) Side (3) Front

Question 4.
For the soldis given below sketch the front, side and top view:
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 15 Visualising Solid Shapes Ex 15.3 Q4.1
Solution:
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 15 Visualising Solid Shapes Ex 15.3 Q4.2

ML Aggarwal Class 7 Solutions for ICSE Maths

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 15 Visualising Solid Shapes Ex 15.2

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 15 Visualising Solid Shapes Ex 15.2

Question 1.
The dimensions of a cuboid are 5 cm, 3 cm, and 2 cm. Draw three different isometric sketches of this cuboid with height
(i) 2 cm
(ii) 3 cm
(iii) 5 cm
Solution:
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 15 Visualising Solid Shapes Ex 15.2 Q1.1

Question 2.
Give (i) an oblique sketch (ii) an isometric sketch for each of the following:
(a) A cube is with an edge 4 cm long.
(b) A cuboid of length 6 cm, breadth 4 cm, and height 3 cm.
Solution:
(a) Cube
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 15 Visualising Solid Shapes Ex 15.2 Q2.1
(b) Cuboid
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 15 Visualising Solid Shapes Ex 15.2 Q2.2

Question 3.
The adjoining figure shows an isometric sketch of a 3-D shape. Draw an oblique sketch that matches this sketch.
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 15 Visualising Solid Shapes Ex 15.2 Q3.1
Solution:
The oblique sketch of the given isometric sketch
of a 3-D shape has been given below:
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 15 Visualising Solid Shapes Ex 15.2 Q3.2

Question 4.
Draw an isometric sketch of each of the following oblique shapes:
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 15 Visualising Solid Shapes Ex 15.2 Q4.1
Solution:
Isometric sketch of each of the given oblique shapes
have been given below:
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 15 Visualising Solid Shapes Ex 15.2 Q4.2

ML Aggarwal Class 7 Solutions for ICSE Maths

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 15 Visualising Solid Shapes Ex 15.1

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 15 Visualising Solid Shapes Ex 15.1

Question 1.
Match the following shapes with their names:
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 15 Visualising Solid Shapes Ex 15.1 Q1.1
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 15 Visualising Solid Shapes Ex 15.1 Q1.2
Solution:
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 15 Visualising Solid Shapes Ex 15.1 Q1.3
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 15 Visualising Solid Shapes Ex 15.1 Q1.4

Question 2.
Identify the nets which can be folded to form a cube (cut out copies of the nets and try it):
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 15 Visualising Solid Shapes Ex 15.1 Q2.1
Solution:
Nets (ii), (in), (iv) and (vi) form a cube.

Question 3.
Dice are cubes with dots on each face. Opposite faces of a die always have a total of sevend dots on them.
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 15 Visualising Solid Shapes Ex 15.1 Q3.1
Here are two nets to make dice (cubes); the numbers inserted in each square indicate the number of dots in that box.
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 15 Visualising Solid Shapes Ex 15.1 Q3.2
Insert suitable numbers in the blanks, remembering that the number on the opposite faces should total to 7.
Solution:
The suitable numbers have been inserted in the boxes.
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 15 Visualising Solid Shapes Ex 15.1 Q3.3

Question 4.
Can any of the following be a net for a die? If no, explain your answer:
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 15 Visualising Solid Shapes Ex 15.1 Q4.1
Solution:
(i) Yes.
(ii) No, because one pair of opposite faces will have
1 and 4 on them whose total not 7, and another pair of
opposite faces will have 3 and 6 on them whose total is also not 7.

Question 5.
Here is an incomplete net for making a cube. Complete it in at least two different ways. Remember that a cube has six faces. How many are there in the net here? (You may use a squared paper for easy manipulation.)
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 15 Visualising Solid Shapes Ex 15.1 Q5.1
Solution:
The given incomplete net of a cube has been
completed in two ways as given under:
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 15 Visualising Solid Shapes Ex 15.1 Q5.2
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 15 Visualising Solid Shapes Ex 15.1 Q5.3
(iii) Three faces out of 6 faces are given.

ML Aggarwal Class 7 Solutions for ICSE Maths

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 14 Symmetry Check Your Progress

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 14 Symmetry Check Your Progress

Question 1.
Draw the line (or lines) of symmetry, if any of the following shapes and count their number.
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 14 Symmetry Check Your Progress Q1.1
Solution:
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 14 Symmetry Check Your Progress Q1.2

Question 2.
For each of the given shape in Question 1, find the order of the rotational symmetry (If any).
Solution:
(i) 4
(ii) 1
(iii) 2
(iv) 4
(v) 5
(vi) 2
(vii) 1
(viii) 1
(ix) 1

Question 3.
Give the order of rotational symmetry of each of the following figures:
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 14 Symmetry Check Your Progress Q3.1
Solution:
(i) 4
(ii) 3
(iii) 8
(iv) 8
(v) 2
(vi) 5

ML Aggarwal Class 7 Solutions for ICSE Maths

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 14 Symmetry Objective Type Questions

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 14 Symmetry Objective Type Questions

Mental Maths

Question 1.
Fill in the blanks:
(i) The hands of a clock rotate only in ………. direction.
(ii) While opening the cap of a bottle, the direction of rotation is ……….
(iii) The number of lines of symmetry of an isosceles right-angled triangle is …………
(iv) A figure has ……….. symmetry if it is its own image under a reflection.
(v) The centre of rotation of an equilateral triangle is the point of intersection of its ………..
(vi) The centre of rotation of a rhombus is the point ……….
(vii) A regular polygon of n-sides has ……….. the number of lines of symmetry.
(viii) The angle of rotational symmetry in an equilateral triangle is ……….
(ix) The angle of rotational symmetry in a regular pentagon is ……..
(x) If after a rotation of 45° about a fixed point the figure looks exactly the same, then the order of rotational symmetry is ……….
Solution:
(i) The hands of a clock rotate only in a clockwise direction.
(ii) While opening the cap of a bottle, the direction of rotation is anticlockwise.
(iii) The number of lines of symmetry of an isosceles right-angled triangle is one.
(iv) A figure has line symmetry if it is its own image under a reflection.
(v) The centre of rotation of an equilateral triangle is
the point of intersection of its medians/angle bisector/altitude.
(vi) The centre of rotation of a rhombus is the point intersection of its diagonals.
(vii) A regular polygon of n-sides has n number of lines of symmetry.
(viii) Angle of rotational symmetry in an equilateral triangle is \(\frac { 360 }{ 3 }\) = 120°.
(ix) Angle of rotational symmetry in a regular pentagon is \(\frac { 360 }{ 5 }\) = 72°.
(x) If after a rotation of 45° about a fixed point the figure looks exactly the same,
then the order of rotational symmetry is \(\frac { 360 }{ 45 }\) = 8.

Question 2.
State whether the following statements are true (T) or false (F):
(i) The letter A has line symmetry but no rotational symmetry.
(ii) A rhombus is also a parallelogram and hence it does not have line symmetry.
(iii) A parallelogram has two lines of symmetry.
(iv) The order of rotational symmetry of a rhombus is four.
(v) A circle has exactly four lines of symmetry.
(vi) In a regular pentagon, the perpendicular bisector of the sides are the only lines of symmetry.
(vii) In a regular hexagon, the perpendicular bisector of the sides are the only lines of symmetry.
(viii) In a rectangle, the angle of rotational symmetry is 90°.
(ix) A semicircle has rotational symmetry of order 2.
(x) An isosceles triangle has neither a line symmetry nor a rotational symmetry.
(xi) If a figure possesses a rotational symmetry, then it must look exactly the same atleast once up to a rotation of 180°.
(xii) The angle of rotation of a figure is obtained by dividing 360° by the order of rotational symmetry.
(xiii) A regular triangle has 3 lines of symmetry and rotational symmetry of order 3.
(xiv) A regular pentagon has 5 lines of symmetry and rotational symmetry of order 5.
Solution:
(i) The letter A has line symmetry but no rotational symmetry. (True)
(ii) A rhombus is also a parallelogram
and hence it does not have line symmetry. (False)
Correct:
As rhombus is a special type of ||gm which has all sides equal.
(iii) A parallelogram has two lines of symmetry. (False)
Correct:
It does not have any line of symmetry.
(iv) The order of rotational symmetry of a rhombus is four. (False)
Correct:
it has two order of rotational symmetry.
(v) A circle has exactly four lines of symmetry. (False)
Correct:
As a circle has infinite lines of symmetry.
(vi) In a regular pentagon,
the perpendicular bisector of the sides are the only lines of symmetry. (True)
(vii) In a regular hexagon,
the perpendicular bisector of the sides are the only lines of symmetry. (False)
Correct:
Its diagonals are also the lines of symmetry.
(viii)In a rectangle, the angle of rotational symmetry is 90°. (False)
Correct:
Its angles are 180°
(ix) A semicircle has rotational symmetry of order 2. (False)
Correct:
It has no order of rotational symmetry.
(x) An isosceles triangle has neither a line symmetry
nor a rotational symmetry. (False)
Correct:
It has one line of symmetry but no rotational symmetry.
(xi) If a figure possesses a rotational symmetry,
then it must look exactly the same at least once up to a rotation of 180°. (True)
(xii) The angle of rotation of a figure is obtained
by dividing 360° by the order of rotational symmetry. (True)
(xiii) A regular triangle has 3 lines of symmetry
and rotational symmetry of order 3. (True)
(xiv) A regular pentagon has 5 lines of symmetry
and rotational symmetry of order 5. (True)

Multiple Choice Questions

Choose the correct answer from the given four options (3 to 14):
Question 3.
A quadrilateral having four lines of symmetry is a
(a) parallelogram
(b) rectangle
(c) rhombus
(d) square
Solution:
A quadrilateral having four lines of symmetry is a square. (d)

Question 4.
The letter Z has
(a) one horizontal line of symmetry
(b) one vertical line of symmetry
(c) two lines of symmetry
(d) no line of symmetry
Solution:
The letter Z has no line of symmetry. (d)

Question 5.
A figure that does not have any rotational symmetry is
(a) circle
(b) parallelogram
(c) kite
(d) regular pentagon
Solution:
A figure that does not have any rotational symmetry is a kite. (c)

Question 6.
The number of lines of symmetry in the given figure is
(a) 1
(b) 3
(c) 6
(d) infinitely many
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 14 Symmetry Objective Type Questions Q6.1
Solution:
The number of lines of symmetry in the given figure is 3. (b)

Question 7.
Rotating a figure by 60° anticlockwise is equivalent to a clockwise rotation of
(a) 60°
(b) 120°
(c) 240°
(d) 300°
Solution:
Rotating a figure by 60° anti-clockwise is equivalent
to a clockwise rotation of 360° – 60° = 300° (d)

Question 8.
A figure having 1 line of symmetry and whose order of rotational symmetry is also 1 is
(a) rhombus
(b) parallelogram
(c) kite
(d) scalene triangle
Solution:
A figure having 1 line of symmetry and
whose order of rotational symmetry is also 1 is a kite. (c)

Question 9.
The order of rotational symmetry of a line segment is
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
The order of rotational symmetry of a line segment is 2. (b)

Question 10.
The order of the rotational symmetry in the given figure is
(a) 1
(b) 2
(c) 4
(d) infinitely many
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 14 Symmetry Objective Type Questions Q10.1
Solution:
The order of the rotational symmetry in the given figure is 2. (b)

Question 11.
A possible angle of rotation of a figure having rotational symmetry of order greater than 1 is
(a) 36°
(b) 144°
(c) 150°
(d) 360°
Solution:
A possible angle of rotation of a figure having
rotational symmetry of order greater than 1 is 360°. (d)

Question 12.
The figure which does not have both reflection and rotational symmetry is
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 14 Symmetry Objective Type Questions Q12.1
Solution:
The figure which does not have
both reflection and rotational symmetry is (c). (c)

Question 13.
In the word ’MATHS’ which of the following pairs of letters have rotational symmetry?
(a) M and T
(b) A and S
(c) T and S
(d) H and S
Solution:
The word ’MATHS’ which of the following
pairs of letters have rotational symmetry H and S. (d)

Question 14.
The letter which has both reflection and rotational symmetry is
(a) H
(b) M
(c) S
(d) Y
Solution:
The letter which has both reflection
and rotational symmetry is H. (a)

ML Aggarwal Class 7 Solutions for ICSE Maths

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 14 Symmetry Ex 14.2

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 14 Symmetry Ex 14.2

Question 1.
Which of the following figures have rotational symmetry? In case of rotational symmetry, find the order of rotational symmetry.
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 14 Symmetry Ex 14.2 Q1.1
Solution:
(i) In figure (i) the rotational symmetry is of order is 2.
(ii) In figure (ii) the rotational symmetry is of order 2.
(iii) In figure (iii) there is no rotational symmetry.
(iv) In figure (iv) the rotational symmetry is of order 2.
(v) In figure (v) there is no rotational symmetry.
(vi) In figure (vi) there is rotational symmetry of order 4.
(vii) In figure (vii) there is rotational symmetry of order 1.
(viii)In figure (viii) there is no rotational symmetry.
(ix) In figure (ix) there is rotational symmetry of order 2.
(x) In figure (x) there is rotational symmetry of order 4.
(xi) In figure (xi) there is rotational symmetry of order 6.
(xii) In figure (xii) there is rotational symmetry of order 4.

Question 2.
Which of the following figures have rotational symmetry of order greater than 1?
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 14 Symmetry Ex 14.2 Q2.1
Solution:
From the given figure,
Figure (i) and (iv) i.e., rhombus and circle have
rotational symmetry more than order 1.
A rhombus has 2 and a circle has many.

Question 3.
Name any two figures that have both lines of symmetry and rotational symmetry.
Solution:
Rhombus and an equilateral triangle have
both line of symmetry and rotational symmetry.

Question 4.
Name the quadrilaterals which have both line and rotational symmetry of order more than 1.
Solution:
In a quadrilateral, rectangle, square and rhombus have both line of symmetry
as well as rotational symmetry.
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 14 Symmetry Ex 14.2 Q4.1

Question 5.
Draw a rough sketch of:
(i) a triangle with both line and rotational symmetries of order more than 1.
(ii) a triangle with only one line of symmetry and no rotational symmetry of order more than 1.
(iii) a triangle with no line symmetry but rotational symmetry of order 1.
(iv) a quadrilateral with no line symmetry but rotational symmetry of order more than 1.
(v) a quadrilateral with line symmetry but not rotational symmetry of order more than 1.
Solution:
(i) A triangle with both line and
rotational symmetry of order more than 1.
It is an equilateral triangle.
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 14 Symmetry Ex 14.2 Q5.1
(ii) A triangle with only one line of symmetry
but no rotational symmetry of order more than 1.
It is an isosceles triangle.
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 14 Symmetry Ex 14.2 Q5.2
(iii) A triangle having no line symmetry
but rotational symmetry of order 1.
It is a scalene triangle.
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 14 Symmetry Ex 14.2 Q5.3
(iv) A quadrilateral with no line symmetry
but rotational symmetry of order more than 1.
It is a parallelogram.
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 14 Symmetry Ex 14.2 Q5.4
(v) A quadrilateral, with the line of symmetry
but not rotational symmetry of order more than 1.
It is an isosceles trapezium.
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 14 Symmetry Ex 14.2 Q5.5

Question 6.
If a figure has two or more than two lines of symmetry, can it have rotational symmetry of order more than one?
Solution:
If it has two or more lines of symmetry, then,
yes, it can have rotational symmetry of order more than one.
A circle is its example.

Question 7.
A figure looks exactly the same as its original figure after rotation of 60°. At what other angles will this figure appear the same?
What can you say if the angle of rotation is
(i) 72°
(ii) 45°
(iii) 50°?
Solution:
A figure looks exactly the same as its original after rotation of 60°.
It will also like the same after rotation of 120°, 180°, 240°, 300°, and 360°.
(i) If the angle of rotation of symmetry is 72°
then it will look exactly the same after it rotates after 144°, 216°, 288°, 360°.
(ii) If the angle of rotation of symmetry is 45°,
then it will look exactly the same after 90°, 135°, 180°, 225°, 270°, 315°, and 360°.
(iii) If the angle of rotation is 50°, then it is not possible.

Question 8.
Can a figure possessing rotational symmetry have an angle of rotation of measure
(i) 180°
(ii) 120°
(iii) 90°
(iv) 30°
(v) 15°
(vi) 17°?
Solution:
(i) 180° : Yes, it can have as \(\frac { 360 }{ 180 }\) = 2°
(ii) 120° : Yes, it can have as \(\frac { 360 }{ 120 }\) = 3°
(iii) 90° : Yes, it can have as \(\frac { 360 }{ 90 }\) = 4°
(iv) 30° : Yes, it can have as \(\frac { 360 }{ 30 }\) = 12°
(v) 15° : Yes, it can have as \(\frac { 360 }{ 15 }\)= 24°
(vi) 17° : No, it cannot have as \(\frac { 360 }{ 17 }\) is not exactly divisible.

ML Aggarwal Class 7 Solutions for ICSE Maths

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 14 Symmetry Ex 14.1

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 14 Symmetry Ex 14.1

Question 1.
Draw all lines of symmetry, if any, in each of the following figures:
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 14 Symmetry Ex 14.1 Q1.1
Solution:
The line/lines of symmetry have been drawn as given below:
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 14 Symmetry Ex 14.1 Q1.2

Question 2.
Copy the figures with a punched hole(s) and draw all the axes of symmetry in each of the following:
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 14 Symmetry Ex 14.1 Q2.1
Solution:
The line/lines of symmetry have been drawn as given below:
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 14 Symmetry Ex 14.1 Q2.2

Question 3.
In the following figure, mark the missing hole(s) in order to make them symmetrical about the dotted line:
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 14 Symmetry Ex 14.1 Q3.1
Solution:
The lines of symmetry have been drawn and the required holes are
marked by dark punches (small circles) as given below:
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 14 Symmetry Ex 14.1 Q3.2

Question 4.
In the following figures, the mirror line (line of symmetry) is given as dotted line. Complete each figure by performing reflection in the mirror (dotted) line and name the figure you complete:
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 14 Symmetry Ex 14.1 Q4.1
Solution:
Each figure is given, has been completed along with the mirror (dotted) line:
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 14 Symmetry Ex 14.1 Q4.2

Question 5.
Copy the adjoining figure.
Take any one diagonal as a line of symmetry and shade a few more squares to make the figure symmetric about a diagonal. Is there more than one way to do that? Will the figure be symmetric about both the diagonals?
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 14 Symmetry Ex 14.1 Q5.1
Solution:
We get the same figure if we shade according to the
other diagonal as a line of symmetry.
Also, we get the same figure of we shade
by taking the line joining the mid-point of the opposite sides.
Yes, the figure is symmetrical about both diagonals.
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 14 Symmetry Ex 14.1 Q5.2

Question 6.
Draw the reflection of the following figures/letter in the given mirror line shown dotted:
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 14 Symmetry Ex 14.1 Q6.1
Solution:
The reflection of the given figure/letter in the given mirror line shown
dotted have been drawn as given below:
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 14 Symmetry Ex 14.1 Q6.2

Question 7.
What other names can you give to the line of symmetry of
(i) an isosceles triangle
(ii) rhombus
(iii) circle?
Solution:
(i) An isosceles triangle: We can be called the line of symmetry
as the angle bisector or median of the triangle.
(ii) Rhombus: The lines of symmetry of the rhombus are
also called as the diagonals of the rhombus as they bisect each other at right angles.
(iii) Circle: The lines of symmetry of a circle are also called the diameters of the circle.
As the diameter of a circle is infinite, so the lines of symmetry of a circle are also infinite.

ML Aggarwal Class 7 Solutions for ICSE Maths

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 13 Practical Geometry Check Your Progress

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 13 Practical Geometry Check Your Progress

Question 1.
State whether the following statements are true or false. Justify your answer.
(i) A triangle with lengths of sides 2.5 cm, 3 cm, and 6 cm can be constructed.
(ii) A triangle DEF with EF = 7.2 cm, m∠E = 110° and m∠F = 80° can be constructed.
(iii) If the measure of an acute angle and the length of the hypotenuse of a right-angled triangle are given, then the triangle can be constructed.
Solution:
(i) We know that in a triangle,
the sum of its any two sides is greater than its third side.
Therefore, a triangle with sides 2.5 cm, 3 cm and 6 cm.
2.5 + 3 = 5.5 < 6 cm
This triangle is not possible.
(ii) In triangle DEF with side EF = 7.2 cm
and m∠E =110° and m∠E = 80°.
The sum of these two angles is 110° + 80° = 190°
which is not possible as a triangle has a sum of 180°.
(iii) If the measure of an acute angle and length of the
hypotenuse in a right-angled triangle is given.
Yes this triangle can be constructed,
Measure of third acute angle = 180° – 90° one acute angle
We are given one side i.e., hypotenuse and its ends angles
which is known as ASA criterion.

Question 2.
Draw a line AB and take a point C outside it. Through C, draw a line parallel to AB by using the concept of equal corresponding angles.
Solution:
Steps of construction:

  1. Draw a line AB and take a point C outside it.
  2. Take a point P on AB and join PC.
  3. Construct ∠PCD = ∠CPB and produce C to D and C to E.
    Then, DE is parallel to AB.
    ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 13 Practical Geometry Check Your Progress Q2.1

Question 3.
Draw a triangle PQR with PQ = 4 cm, QR = 3.5 cm and PR = 4 cm. What type of triangle is this?
Solution:
Steps of construction:

  1. Draw a line segment QR = 3.5 cm.
  2. At Q and R as centers and radius 4 cm,
    draw arcs intersecting each other at P.
  3. Join PQ and PR.
    ∆PQR is the required triangle.
    ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 13 Practical Geometry Check Your Progress Q3.1

Question 4.
Construct a triangle ABC with BC = 7.5 cm, AC = 5 cm and m∠C = 60° by using ruler and compasses only.
Solution:
Steps of construction :

  1. Draw a line segment BC = 7.5 cm.
  2. At C, draw a ray CX making an angle of 60°
    and cut off CA = 5 cm.
  3. Join AB.
    ∆ABC is the required triangle.
    ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 13 Practical Geometry Check Your Progress Q4.1

Question 5.
Construct a triangle ABC, given m∠A = 60°, m∠B = 30° and AB = 5.8 cm by using ruler and compasses only.
Solution:
Steps of construction:

  1. Draw a line segment AB = 5.8 cm.
  2. At A draw a ray AX making an angle of 60° and at B,
    a ray BY making an angle of 30° which intersect each other at C.
    ∆ABC is the required triangle.
    ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 13 Practical Geometry Check Your Progress Q5.1

Question 6.
Construct an isosceles right angled triangle ABC, with m∠ABC = 90° and AC = 6 cm.
Solution:
Steps of construction:

  1. Draw a line segment AC = 6 cm.
  2. Draw its perpendicular bisector which intersects AC at O.
  3. With centre O and AC as diameter, draw a semicircle.
  4. The perpendicular bisects intersect the semicircle at B.
  5. Join BA and BC.
    ∆ABC is the required triangle.
    ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 13 Practical Geometry Check Your Progress Q6.1

ML Aggarwal Class 7 Solutions for ICSE Maths

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 13 Practical Geometry Objective Type Questions

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 13 Practical Geometry Objective Type Questions

Higher Order Thinking Skills (HOTS)

Question 1.
Construct a triangle ABC such that BC = 5.2 cm, AB = 4.8 cm and median CM = 3.6 cm.
Solution:
Steps of construction:

  1. Draw a line segment BC = 5.2 cm.
  2. With centre B and radius \(\frac { 4.8 }{ 2 }\) = 2.4 cm and with centre C and radius 3.6 cm,
    draw arcs intersecting each other at E.
  3. Join CE and BE.
  4. Produce BE to A such that EA = BE = 2.4 cm.
  5. Join AC.
    ∆ABC is the required triangle.

    1. ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 13 Practical Geometry Objective Type Questions Q1.1

Question 2.
Construct an isosceles right-angled triangle ABC such that its hypotenuse BC = 6 cm.
Solution:
Steps of construction:

  1. Draw a line segment BC = 6 cm.
  2. Bisect BC at O.
  3. With centre O and BC as diameter draw the same circle.
  4. At O, draw a perpendicular which meets the semicircle at A.
  5. Join AB and AC.
    ∆ABC is the required triangle.

Note: Angle in a semicircle is 90°.
So, ∠A = 90° and AB = AC.
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 13 Practical Geometry Objective Type Questions Q2.1

ML Aggarwal Class 7 Solutions for ICSE Maths

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 13 Practical Geometry Ex 13

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 13 Practical Geometry Ex 13

Question 1.
Draw a line, say l, take a point P outside it. Through P, draw a line parallel to l using ruler and compasses only.
Solution:
Steps of construction:

  1. Draw a line l and take a point P outside it.
  2. Take another point Q on the line l and join PQ.
  3. Construct at P equal to ∠PQR.
    This is the required line which is parallel to l.
    ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 13 Practical Geometry Ex 13 Q1.1

Question 2.
Draw a line l. Draw a perpendicular to l at any point on l. On this perpendicular choose a point A, 3.5 cm away from line l. Through A, draw a line m parallel to l.
Solution:
Steps of construction:

  1. Draw a line l and take a point P on it.
  2. At P draw a ray PQ making an angle of 90°.
    PQ is the required perpendicular on the line l at point P.
    ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 13 Practical Geometry Ex 13 Q2.1

Question 3.
Let l be a line and P be a point not on l. Through P, draw a line m parallel to l. Now join P to any point Q on l. Choose any other point R on m. Through R, draw a line parallel to PQ. If this line meets l at S, then what shape do the two sets of parallel lines inclose?
Solution:
Steps of construction :

  1. P is a line and P is a point not on the line l.
  2. Take point A on it and join PA.
  3. On P, draw an angle equal to ∠PAl and draw a line m which is parallel to l.
  4. Take a point Q and join PQ. From a point R on m,
    draw a line parallel to PQ which meets l at S.
    We see that PQSR is a ||gm.
    ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 13 Practical Geometry Ex 13 Q3.1

Question 4.
Construct a triangle ABC, given that
(i) AB = 5 cm, BC = 6 cm and AC = 7 cm
(ii) AB = 4.5 cm, BC = 5 cm and AC = 6 cm.
Solution:
(i) Steps of construction :

  1. Draw a line segment BC = 6 cm.
  2. With centre B and radius 5 cm and with centre C and radius 7 cm,
    draw arcs which intersect each other at A.
  3. Join AB and AC.
    ∆ABC is the required triangle.
    ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 13 Practical Geometry Ex 13 Q4.1

(ii) Steps of construction :

  1. Draw a line segment BC 5 cm.
  2. With centre B and radius 4.5 cm and with centre C and radius 6 cm,
    draw arcs which intersect each other at A.
  3. Join AB and AC.
    ∆ABC is the required triangle.
    ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 13 Practical Geometry Ex 13 Q4.2

Question 5.
Construct a triangle PQR given that PQ = 5.4 cm, QR = PR = 4.7 cm. Name the triangle.
Solution:
Steps of construction :

  1. Draw a line segment PQ = 5.4 cm.
  2. With centre P and radius 4.7 cm and with centre Q and radius 4.7 cm,
    draw two arcs intersecting each other at R.
  3. Join RP and RQ.
    PQR is the required triangle which is an isosceles triangle.
    ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 13 Practical Geometry Ex 13 Q5.1

Question 6.
Construct a triangle LMN such that the length of each side is 5.3 cm. Name the triangle.
Solution:
Steps of construction :

  1. Draw a line segment MN = 5.4 cm.
  2. With centre M and N and radius 5.4 cm,
    draw two arcs intersecting each other at L.
    ∆LMN is the required triangle which is an equilateral triangle.
    ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 13 Practical Geometry Ex 13 Q6.1

Question 7.
Construct a triangle ABC such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure ∠ABC and name the triangle.
Solution:
Steps of construction :

  1. Draw a line segment BC = 6 cm.
  2. With centre B and radius 2.5 cm and with centre C and radius 6.5 cm,
    draw two arcs intersecting each other at A.
  3. Join AB and AC.
    ∆ABG is the required triangle.
    On measuring ∠ABC it is equal to 90°.
    Therefore, ∆ABC is a right angled triangle.
    ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 13 Practical Geometry Ex 13 Q7.1

Question 8.
Construct a triangle PQR, given that PQ = 3 cm, QR = 5.5 cm and ∠PQR = 60°.
Solution:
Steps of construction :

  1. Draw a line segment QR = 5.5 cm.
  2. At B, draw a ray BX making an angle of 60°
    and cut off PQ = 3 cm.
  3. Join PR.
    ∆PQR is the required triangle.
    ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 13 Practical Geometry Ex 13 Q8.1

Question 9.
Construct ∆DEF such that DE = 5 cm, DF = 3 cm and m∠EDF = 90°.
Solution:
Steps of construction :

  1. Draw a line segment DE = 5 cm.
  2. At D, draw a ray DX making an angle of 90°
    and cut off DF = 3 cm.
  3. Join FE.
    ∆DEF is the required triangle.
    ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 13 Practical Geometry Ex 13 Q9.1

Question 10.
Construct an isosceles triangle in which the length of each of its equal sides is 6.5 cm and the angle between them is 110°. Measure base angles.
Solution:
Steps of construction:

  1. Draw a line segment AB = 6.5 cm.
  2. At A, draw a ray AX making an angle of 110° and cut off AC = 6.5 cm.
  3. Join BC.
    ∆ABC is the required triangle.
    On measuring its base angles ∠B and ∠C, these are 35° each.
    ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 13 Practical Geometry Ex 13 Q10.1

Question 11.
Construct triangle XYZ if it is given that XY = 6 cm, ∠X = 30° and ∠Y = 100°.
Solution:
Steps of construction :

  1. Draw a line segment XY = 6 cm.
  2. At X, draw a ray XA making an angle of 30° and at Y
    draw a ray YB making an angle of 100°
    which intersect each other at X.
    ∠XYZ is the required triangle.
    ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 13 Practical Geometry Ex 13 Q11.1

Question 12.
Construct a triangle PQR given that PQ = 4.9 cm, ∠P = 45° and ∠Q = 60°. Measure ∠R.
Solution:
Steps of construction :

  1. Draw a line segment PQ = 4.9 cm.
  2. AP, draw a ray PX making are the angle of 45° and at Q,
    draw a ray QY making an angle of 60°
    which intersect it each other at R.
    ∆PQR is the required triangle.
    ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 13 Practical Geometry Ex 13 Q12.1

Question 13.
Construct a triangle ABC such that AB = 4.1 cm, ∠B = 90° and hypotenuse AC = 5.2 cm.
Solution:
Steps of construction:

  1. Draw a line segment AB = 4.1 cm.
  2. AB, draw a ray BX making an angle of 90°.
  3. With centre A and radius 5.2 cm,
    draw an arc which intersects BX at C.
  4. Join AC.
    ∆ABC is the required triangle.
    ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 13 Practical Geometry Ex 13 Q13.1

Question 14.
Construct a right-angled triangle whose hypotenuse is 6 cm long and one of the legs is 4 cm long.
Solution:
Steps of construction :

  1. Draw a line segment BC = 4 cm.
  2. At B, draw a ray BX making an angle of 90°.
  3. With centre C and radius 6 cm
    draw an arc which intersects BX at A.
  4. Join AC.
    ∆ABC is the required triangle.
    ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 13 Practical Geometry Ex 13 Q14.1

ML Aggarwal Class 7 Solutions for ICSE Maths

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 12 Congruence of Triangles Check Your Progress

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 12 Congruence of Triangles Check Your Progress

Question 1.
State, giving reasons, whether the following pairs of triangles are congruent or not:
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 12 Congruence of Triangles Check Your Progress Q1.1
Solution:
(i) In the given figure, using the SSS criterion triangles one congruent.
(ii) Triangles are congruent for the criterion ASA criterion.
(iii) Triangles are congruent for the criterion RHS.
(iv) In the first triangle, third angle = 180° – (70° + 50°) = 180° – 120° = 60°
Now triangles are congruent for ASA criterion.
(v) Not congruent as included angles of the given two sides are not equal.
(vi) Not congruent as the included sides are different.

Question 2.
Given below are measurements of some parts of two triangles. Examine whether the two triangles are congruent or not. In case of congruence, give reasons and write in symbolic form:

∆ABC ∆PQR
(i) AB = 4 cm, BC = 5 cm, ∠B = 70° (i) QR = 4 cm, RP = 5 cm, ∠R = 70°
(ii) AB = 4 cm, BC = 5 cm, ∠B = 80° (ii) PQ = 4 cm, RP = 5 cm, ∠R = 80°
(iii) BC = 6 cm, ∠A = 90°, ∠C = 50° (iii) QR = 6 cm, ∠R = 50°, ZQ = 40°
(iv) AB = 5 cm, ∠A = 90°, BC = 8 cm (iv) PR = 5 cm, ∠P = 90°, QR = 8 cm

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 12 Congruence of Triangles Check Your Progress Q2.1
Solution:
(i) In ∆ABC and ∆PQR
AB = QR = 4 cm
BC = RP = 5 cm
∠B = ∠R = 70°
∆ABC = ∆PQR (SAS criterion)
(ii) In ∆ABC and ∆PQR
AB = PQ = 4 cm
BC = RP = 5 cm not corresponding sides
∠B = ∠R = 80° not corresponding angles
Triangles are not congruent.
(iii) BC = QR = 6 cm
∠A = ∠P = 90° (Third angle)
∠C = ∠R = 50°
Triangles are congruent for ASA criterion.
(iv) AB = PR = 5 cm (Side)
∠A = ∠P = 90°
BC = QR = 8 cm (Hypotenuse)
Triangles are congruent for RHS criterion.

Question 3.
In the given figure, ABC is an isosceles triangle with AB = AC and AD is one of its altitudes.
(i) State the three pairs of equal parts in ∆ADB and ∆ADC.
(zz) Is ∆ADB = ∆ADC? Give reasons.
(iii) Is ∠B = ∠C? Why?
(iv) Is BD = DC? Why?
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 12 Congruence of Triangles Check Your Progress Q3.1
Solution:
∆ABC is an isosceles triangle with AB = AC
and AD is one of the altitudes.
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 12 Congruence of Triangles Check Your Progress Q3.2
(i) In ∆ADB and ∆ADC
Side AD = AD (Common)
Hypotenuse, AB = AC (Given)
∠ADB = ∠ADC = 90° (∵ AB ⊥ BC)
∆ADB = ∆ADC
∠B = ∠C (c.p.c.t)
and BD = CD (c.p.c.t)

Question 4.
In the given figure, OA bisects ∠A and ∠ABO = ∠OCA. Prove that OB = OC.
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 12 Congruence of Triangles Check Your Progress Q4.1
Solution:
In ∆OAB and ∆OAC
∠OAB = ∠OAC (∵ OA bisects ∠A)
∠ABO = ∠ACO (Given)
OA = OA (common)
∆OAB = ∆OAC (AAS congruence rule)
OB = OC (Corresponding parts of congruent As)

Question 5.
In the given figure , prove that
(i) AB = FC
(ii) AF = BC.
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 12 Congruence of Triangles Check Your Progress Q5.1
Solution:
In ∆ABE and ∆DFC
∠B = ∠F (each 90°)
AE = DC (Given)
BE = DF (Given)
∆ABE = ∆DFC (RHS congruence rule)
(i) AB = FC (Corresponding parts of congruent ∆s)
(ii) As AB = FC (Proved above)
⇒ AF + FB = FB + BC
⇒ AF + FB – FB = BC
⇒ AF = BC

ML Aggarwal Class 7 Solutions for ICSE Maths

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 12 Congruence of Triangles Objective Type Questions

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 12 Congruence of Triangles Objective Type Questions

Mental Maths

Question 1.
Fill in the blanks:
(i) Two line segments are congruent if ……….
(ii) Among two congruent angles, one has a measure of 63°; the measure of the other angle is ……….
(iii) When we write ∠A = ∠B, we actually mean ………
(iv) The side included between ∠M and ∠N of ∆MNP is ……….
(v) The side QR of ∆PQR is included between angles ……….
(vi) If two triangles ABC and PQR are congruent under the correspondence A ↔ R, B ↔ P and C ↔ Q, then in symbolic form it can be written as ∆ABC = ………
(vii) If ∆DEF = ∆SRT, then the correspondence between vertices is ……….
Solution:
(i) Two line segments are congruent if they are of the same length.
(ii) Among two congruent angles, one has a measure of 63°;
the measure of the other angle is 63°.
(iii) When we write ∠A = ∠B, we actually mean m∠A = m∠B.
(iv) The side included between ∠M and ∠N of ∆MNP is MN.
(v) The side QR of ∆PQR is included between angles ∠Q and ∠R.
(vi) If two triangles ABC and PQR are congruent
under the correspondence A ↔ R, B ↔ P and C ↔ Q,
then in symbolic form it can be written as ∆ABC = ∆RPQ.
(vii) If ∆DEF = ∆SRT, then the correspondence between vertices is
D ↔ S, E ↔ R and F ↔ T.

Question 2.
State whether the following statements are true (T) or false (F):
(i) All circles are congruent.
(ii) Circles having equal radii are congruent.
(iii) Two congruent triangles have equal areas and equal perimeters.
(iv) Two triangles having equal areas are congruent.
(v) Two squares having equal areas are congruent.
(vi) Two rectangles having equal areas are congruent.
(vii) All acute angles are congruent.
(vii)All right angles are congruent.
(ix) Two figures are congruent if they have the same shape.
(x) A two rupee coin is congruent to a five rupee coin.
(xi) All equilateral triangles are congruent.
(xii) Two equilateral triangles having equal perimeters are congruent.
(xii) If two legs of one right triangle are equal to two legs of another right angle triangle, then the two triangles are congruent by SAS rule.
(xiv) If three angles of two triangles are equal, then triangles are congruent.
(xv) If two sides and one angle of one triangle are equal to two sides and one angle of another triangle, then the triangle are congruent.
Solution:
(i) All circles are congruent. (False)
Correct:
As if all circles have equal radii otherwise not.
(ii) Circles having equal radii are congruent. (True)
(iii) Two congruent triangles have equal areas
and equal perimeters. (True)
(iv) Two triangles having equal areas are congruent. (False)
Correct:
As they may have different sides and angles.
(v) Two squares having equal areas are congruent. (True)
(vi) Two rectangles having equal areas are congruent. (False)
Correct:
As their side can be different.
(vii) All acute angles are congruent. (False)
Correct:
As acute angles have different measures.
(viii) All right angles are congruent. (True)
(ix) Two figures are congruent if they have the same shape. (False)
Correct:
As the same shapes have different measures.
(x) A two rupee coin is congruent to a five rupee coin. (False)
Correct:
As they have different size.
(xi) All equilateral triangles are congruent. (False)
Correct:
As they have different sides in length.
(xii) Two equilateral triangles having equal perimeters are congruent. (True)
(xiii) If two legs of one right triangle are equal to
two legs of another right angle triangle,
then the two triangles are congruent by SAS rule. (True)
(xiv) If three angles of two triangles are equal,
then triangles are congruent. (False)
Correct:
They can be similar to each other.
(xv) If two sides and one angle of one triangle are equal to two sides
and one angle of another triangle, then the triangle is congruent. (False)
Correct:
If the angles are included, they can be congruent.

Multiple Choice Questions

Choose the correct answer from the given four options (3 to 14):
Question 3.
Which one of the following is not a standard criterion of congruency of two triangles?
(a) SSS
(b) SSA
(c) SAS
(d) ASA
Solution:
The axiom SSA is not a standard criterion
of congruency of triangles. (b)

Question 4.
If ∆ABC = ∆PQR and ∠CAB = 65°, then ∠RPQ is
(a) 65°
(b) 75°
(c) 90°
(d) 115°
Solution:
∆ABC = ∆PQR
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 12 Congruence of Triangles Objective Type Questions Q4.1
∠CAB = 65°
∠RPQ = 65° (corresponding angles) (a)

Question 5.
If ∆ABC = ∆EFD, then the correct statement is
(a) ∠A = ∠D
(b) ∠A = ∠F
(c) ∠A = ∠E
(d) ∠B = ∠E
Solution:
∆ABC = ∆EFD
Then ∠A = ∠E (c)
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 12 Congruence of Triangles Objective Type Questions Q5.1

Question 6.
If ∆ABC = ∆PQR, then the correct statement is
(a) AB = QR
(b) AB = PR
(c) BC = PR
(d) AC = PR
Solution:
∆ABC = ∆PQR
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 12 Congruence of Triangles Objective Type Questions Q6.1
Then AB = PQ
AC = PR (d)

Question 7.
If ∠D = ∠P, ∠E = ∠Q and DE = PQ, then ∆DEF = ∆PQR, by the congruence rule
(a) SAS
(b) ASA
(c) SSS
(d) RHS
Solution:
In ∆DEF = ∆PQR
∠D = ∠P, ∠E = ∠Q
DE = PQ
∆DEF = ∆PQR (ASA axiom) (b)

Question 8.
In ∆ABC and ∆PQR, BC = QR and ∠C = ∠R. To establish ∆ABC = ∆PQR by SAS congruence rule, the additional information required is
(a) AC = PR
(b) AB = PR
(c) CA = PQ
(d) AB = PQ
Solution:
If ∆ABC = ∆PQR by SAS
BC = QR and ∠C = ∠R, then AC = PR (a)

Question 9.
In the given figure, the lengths of the sides of two triangles are given. The correct statement is
(a) ∆ABC = ∆PQR
(b) ∆ABC = ∆QRP
(c) ∆ABC = ∆QPR
(d) ∆ABC = ∆RPQ
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 12 Congruence of Triangles Objective Type Questions Q9.1
Solution:
Correct statement is ∆ABC = ∆QRP. (b)

Question 10.
In the given figure, M is the mid-point of both AC and BD. Then
(a) ∠1 = ∠2
(b) ∠1 = ∠4
(c) ∠2 = ∠4
(d) ∠1 = ∠3
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 12 Congruence of Triangles Objective Type Questions Q10.1
Solution:
In the given figure,
M is mid-point of AC and BD both then ∠1 = ∠4. (b)

Question 11.
In the given figure, ∆PQR = ∆STU. What is the length of TU?
(a) 5 cm
(b) 6 cm
(c) 7 cm
(d) cannot be determined
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 12 Congruence of Triangles Objective Type Questions Q11.1
Solution:
In the given figure,
∆PQR = ∆STU
TU = QR = 6 cm (b)

Question 12.
In the given figure, ∆ABC and ∆DBC are on the same base BC. If AB = DC and AC = DB, then which of the following statement is correct?
(a) ∆ABC = ∆DBC
(b) ∆ABC = ∆CBD
(c) ∆ABC = ∆DCB
(d) ∆ABC = ∆BCD
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 12 Congruence of Triangles Objective Type Questions Q12.1
Solution:
In the given figure,
AB = DC, AC = DB
Then, ∆ABC = ∆DCB (c)

Question 13.
The two triangles shown in the given figure are:
(a) congruent by AAS rule
(b) congruent by ASA rule
(c) congruent by SAS rule
(d) not congruent.
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 12 Congruence of Triangles Objective Type Questions Q13.1
Solution:
In the given two triangles are not congruent.
In first triangle, AAS are given while in second ASA are given. (d)

Question 14.
In .the given figure, ∆ABC = ∆PQR. The values of x and y are:
(a) x = 63, y = 35
(b) x = 77, y = 35
(c) x = 35, y = 77
(d) x = 63, y = 40
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 12 Congruence of Triangles Objective Type Questions Q14.1
Solution:
In the given figure,
∆ABC = ∆PQR
∠A = ∠P and ∠B = ∠Q
Now x – 7 = 70°
⇒ x = 70° + 7 = 77°
and 2y + 5 = 75
⇒ 2y = 75° – 5 = 70°
⇒ y = 35°
x = 77°, y = 35° (b)

Higher Order Thinking Skills (HOTS)

Question 1.
If all the three altitudes of a triangle are equal, then prove that it is an equilateral triangle.
Solution:
Given: In ∆ABC,
AD, BE and CF are altitudes of the triangle
and AD = BE = CF.
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 12 Congruence of Triangles Objective Type Questions hotsQ1.1
To prove: ∆ABC is an equilateral.
Proof: In ∆ABD and ∆CFB
AD = CF (Given)
∠D = ∠F (Each = 90°)
∠B = ∠B (Common)
∆ABD = ∆CFB (AAS criterion)
AB = BC …….(i)
Similarly in ∆BEC and ∆ADC
BE = AD (Given)
∠C = ∠C (Common)
∠E = ∠D (Each = 90°)
∆BEC = ∆ADC (AAS criterion)
BC = AC ………(ii)
From (i) and (ii)
AB = BC = AC
∆ABC is an equilateral triangle.

Question 2.
In the given fig., if BA || RP, QP || BC and AQ = CR, then prove that ∆ABC = ∆RPQ.
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 12 Congruence of Triangles Objective Type Questions hotsQ2.1
Solution:
In the given figure, BA || RP
QP || BC and AQ = CR
To prove : ∆ABC = ∆RPQ
Proof: AQ = CR
Adding CQ to both sides
AQ + CQ = CR + CQ
⇒ AC = RQ
Now in ∆ABC and ∆RPQ
∠A = ∠R (Alternate angles)
∠C = ∠Q (Alternate angles)
AC = RQ (Proved)
∆ABC = ∆RPQ (ASA criterion)
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 12 Congruence of Triangles Objective Type Questions hotsQ2.2

ML Aggarwal Class 7 Solutions for ICSE Maths