Samacheer Class 7 Maths Measurements Intext Questions Book Back Answers & Solutions

Q.1Explain the area of the parallelogram as sum of the areas of the two triangles.v

Solution

ABCD is a parallelogram. It can be divided into two triangles of equal area by drawing the diagonal BD.
Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Intext Questions 8
Area of the parallelogram ABCD = base × height
= AB × DE

Answer:

ABCD is a parallelogram. It can be divided into two triangles of equal area by drawing the diagonal BD.
Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Intext Questions 8
Area of the parallelogram ABCD = base × height
= AB × DE

Q.2A rectangle is a parallelogram but a parallelogram is not a rectangle. Why?v

Solution

(i) For both rectangle and parallelogram
(i) opposite sides are equal and parallel.
(ii) For rectangle all angles equal to 90°. But for parallelogram opposite angles are equal.
∴ All rectangles are parallelograms. But all parallelograms are not rectan¬gles as their angles need not be equal to 90°.
(Try These Textbook Page No. 36)

Answer:

(i) For both rectangle and parallelogram
(i) opposite sides are equal and parallel.
(ii) For rectangle all angles equal to 90°. But for parallelogram opposite angles are equal.
∴ All rectangles are parallelograms. But all parallelograms are not rectan¬gles as their angles need not be equal to 90°.
(Try These Textbook Page No. 36)

Q.1Count the squares and find the area of the following parallelograms by converting those into rectangles of the same area. (Without changing the base and height). Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Intext Questions 10

Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Intext Questions 10

v

A. ______ sq. units
B. ______ sq. units
C. ______ sq. units
D. ______ sq. units

Solution

Converting the given parallelograms into rectangles we get.
Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Intext Questions 11
(a) 10 sq. units
(b) 18 sq. units
(c) 16 sq. units
(d) 5 sq. units

Answer:

Converting the given parallelograms into rectangles we get.
Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Intext Questions 11
(a) 10 sq. units
(b) 18 sq. units
(c) 16 sq. units
(d) 5 sq. units

Q.3Find the area o the following parallelograms by measuring their base and height, using formula. Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Intext Questions 18

Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Intext Questions 18

(e) _____ sq. unitsv

A. _____ sq. units
B. _____ sq. units
C. _____ sq. units
D. _____ sq. units

Solution

(a) Area of the rectangle = (base × height) sq. units
base = 5 units
height = 5 units
∴ Area = (5 × 5 ) = sq. units = 25 sq. units
(b) Area of the rectangle = (base × height) sq. units
base = 4 units
height = 1 units
∴ Area = (4 × 1 ) = sq. units = 4 sq. units
(c) Area of the rectangle = (base × height) sq. units
base = 2 units
height = 3 units
∴ Area = (2 × 3 ) = sq. units = 6 sq. units
(d) Area of the rectangle = (base × height) sq. units
base = 4 units
height = 4 units
∴ Area = (4 × 4 ) = sq. units = 16 sq. units
(e) Area of the parallelogram = (base × height) sq. units
base = 7 units
height = 5 units
= 7 × 5 = 35 sq. units
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Answer:

(a) Area of the rectangle = (base × height) sq. units
base = 5 units
height = 5 units
∴ Area = (5 × 5 ) = sq. units = 25 sq. units
(b) Area of the rectangle = (base × height) sq. units
base = 4 units
height = 1 units
∴ Area = (4 × 1 ) = sq. units = 4 sq. units
(c) Area of the rectangle = (base × height) sq. units
base = 2 units
height = 3 units
∴ Area = (2 × 3 ) = sq. units = 6 sq. units
(d) Area of the rectangle = (base × height) sq. units
base = 4 units
height = 4 units
∴ Area = (4 × 4 ) = sq. units = 16 sq. units
(e) Area of the parallelogram = (base × height) sq. units
base = 7 units
height = 5 units
= 7 × 5 = 35 sq. units
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Q.4Draw as many parallelograms as possible in a grid sheet with the area 20 square units each.v

Solution

Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Intext Questions 19
Area of parallelogram (a), (b) or (c) = 20 sq. units
Exercise 2.2
Rhombus
(Try These Textbook Page No. 41)

Answer:

Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Intext Questions 19
Area of parallelogram (a), (b) or (c) = 20 sq. units
Exercise 2.2
Rhombus
(Try These Textbook Page No. 41)

Q.3Can you find the perimeter of the rhombus?v

Solution

If we know the length of one side we can find the perimeter using 4 × side units.

Answer:

If we know the length of one side we can find the perimeter using 4 × side units.

Q.4Can diagonals of a rhombus be of the same length?v

Solution

When the diagonals of a rhombus become equal it become a square.

Answer:

When the diagonals of a rhombus become equal it become a square.

Q.5A square is a rhombus but a rhombus is not a square. Why?v

Solution

In a square
(i) all sides are equal.
(ii) opposite sides are parallel
(iii) diagonals divides the square into 4 right angled triangles of equal area
(iv) the diagonals bisect each other at right angles.
So it become a rhombus also.
But in a rhombus (i) each angle need not equal to 90°.
(ii) the length of the diagonals need not be equal. Therefore it does not become a square.

Answer:

In a square
(i) all sides are equal.
(ii) opposite sides are parallel
(iii) diagonals divides the square into 4 right angled triangles of equal area
(iv) the diagonals bisect each other at right angles.
So it become a rhombus also.
But in a rhombus (i) each angle need not equal to 90°.
(ii) the length of the diagonals need not be equal. Therefore it does not become a square.

Q.6Can you draw a rhombus in such a way that the side is equal to the diagonal.v

Solution

Yes, we can draw a rhombus with one of its diagonals equal to its side length. In such case the diagonal will divide the rhombus into two congruent equilateral triangles.
Exercise 2.3
(Try These Textbook Page No. 46)

Answer:

Yes, we can draw a rhombus with one of its diagonals equal to its side length. In such case the diagonal will divide the rhombus into two congruent equilateral triangles.
Exercise 2.3
(Try These Textbook Page No. 46)

Q.1Can you find the perimeter of the trapezium? Discuss.v

Solution

If all sides are given, then by adding all the four lengths we can find the perimeter of a trapezium.

Answer:

If all sides are given, then by adding all the four lengths we can find the perimeter of a trapezium.

Q.3Mention any three life situations where the isosceles trapeziums are used?v

Solution

(i) Glass of a car windows.
(ii) Eye glass (glass in spectacles)
(iii) Some bridge supports.
(iv) Sides of handbags.
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Answer:

(i) Glass of a car windows.
(ii) Eye glass (glass in spectacles)
(iii) Some bridge supports.
(iv) Sides of handbags.
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Q.1Find the area and perimeter of the following parallelograms. Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.1 1

v

Solution

(i) Given base b = 11 cm ; height h = 3 cm
Area of the parallelogram = b × h sq. units = 11 × 3 cm 2
= 33 cm 2
Also perimeter of a parallelogram = Sum of 4 sides
= 11 cm + 4 cm + 11 cm + 4 cm = 30 cm
Area = 33 cm 2 ; Perimeter = 30 cm.
(ii) Given base b = 7 cm
height h = 10 cm
Area of the parallelogram = b × h sq. units
= 7 × 10 cm 2 = 70 cm 2
Perimeter = Sum of four sides
= 13 cm + 7 cm + 13 cm + 7 cm = 40 cm
Area = 70 cm 2 , Perimeter = 40 cm

Answer:

(i) Given base b = 11 cm ; height h = 3 cm
Area of the parallelogram = b × h sq. units = 11 × 3 cm 2
= 33 cm 2
Also perimeter of a parallelogram = Sum of 4 sides
= 11 cm + 4 cm + 11 cm + 4 cm = 30 cm
Area = 33 cm 2 ; Perimeter = 30 cm.
(ii) Given base b = 7 cm
height h = 10 cm
Area of the parallelogram = b × h sq. units
= 7 × 10 cm 2 = 70 cm 2
Perimeter = Sum of four sides
= 13 cm + 7 cm + 13 cm + 7 cm = 40 cm
Area = 70 cm 2 , Perimeter = 40 cm

Q.3Suresh on a parallelogram shaped trophy in a state level chess tournament. He knows that the area of the trophy is 735 sq. cm and its base is 21 cm. What is the height of that trophy? Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.1 4

v

Solution

Given base 6 = 21 cm
Area of parallelogram = 735 sq. cm
b × h = 735
21 × h = 735
h = \(\frac{735}{21}\)
h = 35 cm
∴ Height of the trophy = 35 cm
SamacheerKalvi.Guru

Answer:

Given base 6 = 21 cm
Area of parallelogram = 735 sq. cm
b × h = 735
21 × h = 735
h = \(\frac{735}{21}\)
h = 35 cm
∴ Height of the trophy = 35 cm
SamacheerKalvi.Guru

Q.4Janaki has a piece of fabric in the shape of a parallelogram. Its height is 12 m and its base is 18 m. She cuts the fabric into four equal parallelograms by cutting the parallel sides through its mid-points. Find the area of each new parallelogram.v

Solution

Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.1 5
Area of a parallelogram = (base × height) sq. units
Base length = \(\frac{18}{2}\) = 9 m
Height = \(\frac{12}{2}\) = 6 m
Area = 9 × 6 = 54 m 2
Area of each parallelogram = 54 m 2

Answer:

Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.1 5
Area of a parallelogram = (base × height) sq. units
Base length = \(\frac{18}{2}\) = 9 m
Height = \(\frac{12}{2}\) = 6 m
Area = 9 × 6 = 54 m 2
Area of each parallelogram = 54 m 2

Q.5A ground is in the shape of parallelogram. The height of the parallelogram is 14 metres and the corresponding base is 8 metres longer than its height. Find the cost of levelling the ground at the rate of ₹ 15 per sq. m.v

Solution

Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.1 6
Height of the parallelogram h = 14 m
Base = 8 m longer than height
= (14 + 8) m = 22 m
Area of the parallelogram = (base × height) sq. units
= (22 × 14)m 2 = 308 m 2
Cost of levelling 1 m 2 = ₹ 15
Cost of levelling 308 m 2 = 308 × 15 = ₹ 4,620
Cost of levelling the ground = ₹ 4,620
Objective Type Questions

Answer:

Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.1 6
Height of the parallelogram h = 14 m
Base = 8 m longer than height
= (14 + 8) m = 22 m
Area of the parallelogram = (base × height) sq. units
= (22 × 14)m 2 = 308 m 2
Cost of levelling 1 m 2 = ₹ 15
Cost of levelling 308 m 2 = 308 × 15 = ₹ 4,620
Cost of levelling the ground = ₹ 4,620
Objective Type Questions

Q.6The perimeter of a parallelogram whose adjacent sides are 6 cm and 5 cm is (i) 12 cm (ii) 10 cm (iii) 24 cm (iv) 22 cmv

Solution

(iv) 22 cm
Hint:
= 2(6 + 5) = 2 × 11 = 22 cm

Answer:

(iv) 22 cm
Hint:
= 2(6 + 5) = 2 × 11 = 22 cm

Q.7The area of parallelogram whose base 10 m and height 7 m is (i) 70 sq.m (ii) 35 sq.m (iii) 7 sq.m (iv) 10 sq.mv

Solution

(i) 70 sq. m
Hint: = base × height = 10m × 7m = 70 sq.m

Answer:

(i) 70 sq. m
Hint: = base × height = 10m × 7m = 70 sq.m

Q.8The base of the parallelogram with area is 52 sq. cm and height 4 cm is (i) 48 cm (ii) 104 cm (iii) 13 cm (iv) 26 cmv

Solution

(iii) 13 cm
Hint:
Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.1 7

Answer:

(iii) 13 cm
Hint:
Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.1 7

Q.9What happens to the area of the parallelogram if the base is increased 2 times and the height is halved? (i) Decreases to half (ii) Remains the same (iii) Increase by two times (iv) Nonev

Solution

(ii) Remains the same
Hint:
Area = b × h sq. units
New base = 2 × old base
New height = \(\frac{1}{2}\) × old height
New Area = New base × New height = (2 × b)\(\frac{1}{2}\) × h = bh = old Area.
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Answer:

(ii) Remains the same
Hint:
Area = b × h sq. units
New base = 2 × old base
New height = \(\frac{1}{2}\) × old height
New Area = New base × New height = (2 × b)\(\frac{1}{2}\) × h = bh = old Area.
SamacheerKalvi.Guru

Q.10In a parallelogram the base is three times its height. If the height is 8 cm then the area is (i) 64 sq. cm (ii) 192 sq. cm (iii) 32 sq. cm (iv) 72 sq. cmv

Solution

(ii) 192 sq. cm
Hint: Given b = 3 × h; h = 8 cm
Area = b × h = 3h × 8 = 3 × 8 × 8 = 192 cm 2
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Answer:

(ii) 192 sq. cm
Hint: Given b = 3 × h; h = 8 cm
Area = b × h = 3h × 8 = 3 × 8 × 8 = 192 cm 2
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Q.1Find the area of rhombus PQRS shown in the following figures. Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.2 1

v

Solution

(i) Given the diagonals d 1 = 16 cm ; d 2 = 8 cm
Area of the rhombus = \(\frac{1}{2}\)(d 1 × d 2 ) sq. units
= \(\frac{1}{2}\) × 16 × 8 cm 2 = 64 cm 2
Area of the rhombus = 64 cm 2
(ii) Given base b = 15 cm ; Height h = 11 cm
Area of the rhombus = (base × height) sq. units
= 15 × 11 cm 2 = 165 cm 2
Area of the rhombus = 165 cm 2

Answer:

(i) Given the diagonals d 1 = 16 cm ; d 2 = 8 cm
Area of the rhombus = \(\frac{1}{2}\)(d 1 × d 2 ) sq. units
= \(\frac{1}{2}\) × 16 × 8 cm 2 = 64 cm 2
Area of the rhombus = 64 cm 2
(ii) Given base b = 15 cm ; Height h = 11 cm
Area of the rhombus = (base × height) sq. units
= 15 × 11 cm 2 = 165 cm 2
Area of the rhombus = 165 cm 2

Q.2Find the area of a rhombus whose base is 14 cm and height is 9 cm.v

Solution

Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.2 2
Given base b = 14 cm ; Height h = 9 cm
Area of the rhombus = b × h sq. units
= 14 × 9 cm 2 = 126 cm 2

Answer:

Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.2 2
Given base b = 14 cm ; Height h = 9 cm
Area of the rhombus = b × h sq. units
= 14 × 9 cm 2 = 126 cm 2

Q.4The area of a rhombus is 100 sq. cm and length of one of its diagonals is 8 cm. Find the length of the other diagonal.v

Solution

Given the length of one diagonal d 1 = 8 cm ; Area of the rhombus = 100 sq. cm
\(\frac{1}{2}\)(d 1 × d 2 ) = 100
\(\frac{1}{2}\) × 8 × d 2 = 100
8 × d 2 = 100 × 2
d 2 = \(\frac{100 \times 2}{8}\) = 25 cm
Length of the other diagonal d 2 = 25 cm
SamacheerKalvi.Guru

Answer:

Given the length of one diagonal d 1 = 8 cm ; Area of the rhombus = 100 sq. cm
\(\frac{1}{2}\)(d 1 × d 2 ) = 100
\(\frac{1}{2}\) × 8 × d 2 = 100
8 × d 2 = 100 × 2
d 2 = \(\frac{100 \times 2}{8}\) = 25 cm
Length of the other diagonal d 2 = 25 cm
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Q.5A sweet is in the shape of rhombus whose diagonals are given as 4 cm and 5 cm. The surface of the sweet should be covered by an aluminum foil. Find the cost of aluminum foil used for 400 such sweets at the rate of ₹ 7 per 100 sq. cm.v

Solution

Diagonals d 1 = 4 cm and d 2 = 5 cm
Area of one rhombus shaped sweet = \(\frac{1}{2}\)(d 1 × d 2 ) sq. units = \(\frac{1}{2}\) × 4× 5 cm 2 = 10 cm 2
Aluminum foil used to cover 1 sweet = 10 cm 2
∴ Aluminum foil used to cover 400 sweets = 400 × 10 = 4000 cm 2
Cost of aluminum foil for 100 cm 2 = ₹ 7
∴ Cost of aluminum foil for 4000 cm 2 = \(\frac{4000}{100}\) × 7 = ₹ 280
∴ Cost of aluminum foil used = ₹ 280.
Objective Type Questions

Answer:

Diagonals d 1 = 4 cm and d 2 = 5 cm
Area of one rhombus shaped sweet = \(\frac{1}{2}\)(d 1 × d 2 ) sq. units = \(\frac{1}{2}\) × 4× 5 cm 2 = 10 cm 2
Aluminum foil used to cover 1 sweet = 10 cm 2
∴ Aluminum foil used to cover 400 sweets = 400 × 10 = 4000 cm 2
Cost of aluminum foil for 100 cm 2 = ₹ 7
∴ Cost of aluminum foil for 4000 cm 2 = \(\frac{4000}{100}\) × 7 = ₹ 280
∴ Cost of aluminum foil used = ₹ 280.
Objective Type Questions

Q.6The area of the rhombus with side 4 cm and height 3 cm is (i) 7 sq. cm (ii) 24 sq. cm (iii) 12 sq. cm (iv) 10 sq. cmv

Solution

(iii) 12 sq. cm
Hint:
Area = Base × Height = 4 × 3 = 12 cm 2

Answer:

(iii) 12 sq. cm
Hint:
Area = Base × Height = 4 × 3 = 12 cm 2

Q.7The area of the rhombus when both diagonals measuring 8 cm is (i) 64 sq. cm (ii) 32 sq. cm (iii) 30 sq. cm (iv) 16 sq. cmv

Solution

(ii) 32 sq. cm
Hint:
Area = \(\frac{1}{2}\)(d 1 × d 2 ) = \(\frac{1}{2}\) × 8 × 8 = 32

Answer:

(ii) 32 sq. cm
Hint:
Area = \(\frac{1}{2}\)(d 1 × d 2 ) = \(\frac{1}{2}\) × 8 × 8 = 32

Q.8The area of the rhombus is 128 sq. cm. and the length of one diagonal is 32 cm. The length of the other diagonal is (i) 12 cm (ii) 8 cm (iii) 4 cm (iv) 20 cmv

Solution

(ii) 8 cm
Hint:
\(\frac{1}{2}\) × d 1 × d 2 = 128 ⇒ d 2 = \(\frac{128 \times 2}{32}\) = 8cm
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Answer:

(ii) 8 cm
Hint:
\(\frac{1}{2}\) × d 1 × d 2 = 128 ⇒ d 2 = \(\frac{128 \times 2}{32}\) = 8cm
SamacheerKalvi.Guru

Q.9The height of the rhombus whose area 96 sq. m and side 24 m is (i) 8 m (ii) 10 m (iii) 2 m (iv) 4 mv

Solution

(iv) 4 m
Hint:
Area = Base × height = 96 ⇒ height = \(\frac{96}{24}\) = 4

Answer:

(iv) 4 m
Hint:
Area = Base × height = 96 ⇒ height = \(\frac{96}{24}\) = 4

Q.10The angle between the diagonals of a rhombus is (i) 120° (ii) 180° (iii) 90° (iv) 100°v

Solution

(iii) 90°
Hint:
Angles of a rhombus bisect at right angles.
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Answer:

(iii) 90°
Hint:
Angles of a rhombus bisect at right angles.
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Q.2Find the area of a trapezium whose parallel sides are 24 cm and 20 cm and the distance between them is 15 cm.v

Solution

Given the parallel sides a = 24 cm; b = 20 cm
Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.3 3
Distance between a and b is ‘h’ = 15 cm
Area of the trapezium = \(\frac{1}{2}\) × h × (a + b) sq. units
= \(\frac{1}{2}\) × 15 × (24 + 20) cm 2
= \(\frac{1}{2}\) × 15 × 44 = 330 cm 2
Area of the trapezium = 330 cm 2

Answer:

Given the parallel sides a = 24 cm; b = 20 cm
Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.3 3
Distance between a and b is ‘h’ = 15 cm
Area of the trapezium = \(\frac{1}{2}\) × h × (a + b) sq. units
= \(\frac{1}{2}\) × 15 × (24 + 20) cm 2
= \(\frac{1}{2}\) × 15 × 44 = 330 cm 2
Area of the trapezium = 330 cm 2

Q.3The area of a trapezium is 1586 sq. cm. The distance between its parallel sides is 26 cm. If one of the parallel sides is 84 cm then find the other side.v

Solution

Given one parallel side = 84 cm. Let the other parallel side be ‘b’ cm.
Distance between a and b is h = 26 cm.
Area of the trapezium = 1586 sq. cm
Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.3 4
∴ The other parallel side = 38 cm.

Answer:

Given one parallel side = 84 cm. Let the other parallel side be ‘b’ cm.
Distance between a and b is h = 26 cm.
Area of the trapezium = 1586 sq. cm
Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.3 4
∴ The other parallel side = 38 cm.

Q.4The area of a trapezium is 1080 sq. cm. If the lengths of its parallel sides are 55.6 cm and 34.4 cm. Find the distance between them.v

Solution

Length of the parallel sides a = 55.6 cm ; b = 34.4 cm
Area of the trapezium = 1080 sq. cm
Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.3 5
Distance between parallel sides = 24 cm.
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Answer:

Length of the parallel sides a = 55.6 cm ; b = 34.4 cm
Area of the trapezium = 1080 sq. cm
Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.3 5
Distance between parallel sides = 24 cm.
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Q.5The area of a trapezium is 180 sq. cm and its height is 9 cm. If one of the parallel sides is longer than the other by 6 cm. Find the length of the parallel sides.v

Solution

Let one of the parallel side be ‘a’ cm. Given one parallel sides is longer than the other by 6 cm.
i.e. b = a + 6 cm Also given height ‘h’ = 9 cm
Area of trapezium = 180 sq. cm
Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.3 6
2a + 6 = 40
2a = 40 – 6 = 34
a = \(\frac{34}{2}\) = 17 cm
b = a + 6= 17 + 6 = 23 cm
∴ The parallel sides are a = 17 cm and b = 23 cm

Answer:

Let one of the parallel side be ‘a’ cm. Given one parallel sides is longer than the other by 6 cm.
i.e. b = a + 6 cm Also given height ‘h’ = 9 cm
Area of trapezium = 180 sq. cm
Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.3 6
2a + 6 = 40
2a = 40 – 6 = 34
a = \(\frac{34}{2}\) = 17 cm
b = a + 6= 17 + 6 = 23 cm
∴ The parallel sides are a = 17 cm and b = 23 cm

Q.6The sunshade of a window is in the form of isoceles trapezium whose parallel sides are 81 cm and 64 cm and the distance between them is 6 cm. Find the cost of painting the surface at the rate of ₹ 2 per sq. cm.v

Solution

Given the parallel sides a = 81 cm ; b = 64 cm
Distance between ‘a’ and ‘b’ is height h = 6 cm
Area of the trapezium = \(\frac{1}{2}\) × h(a + b) sq. units
= \(\frac{1}{2}\) × 6 3 × (81 + 64) = 3 × 145 cm 2 = 435 cm 2
Cost of painting 1 cm 2 = ₹ 2
Cost of painting 435 cm 2 = ₹ 435 × 2 = ₹ 870
Cost of painting = ₹ 870 .

Answer:

Given the parallel sides a = 81 cm ; b = 64 cm
Distance between ‘a’ and ‘b’ is height h = 6 cm
Area of the trapezium = \(\frac{1}{2}\) × h(a + b) sq. units
= \(\frac{1}{2}\) × 6 3 × (81 + 64) = 3 × 145 cm 2 = 435 cm 2
Cost of painting 1 cm 2 = ₹ 2
Cost of painting 435 cm 2 = ₹ 435 × 2 = ₹ 870
Cost of painting = ₹ 870 .

Q.7A window is in the form of trapezium whose parallel sides are 105 cm and 50 cm respectively and the distance between the parallel sides is 60 cm. Find the cost of the glass used to cover the window at the rate of ₹ 15 per 100 sq. cm.v

Solution

Given the parallel sides a = 105 cm ; b = 50 cm ; Height = 60 cm
Area of the trapezium = \(\frac{1}{2}\) × h × (a + b)sq. units = \(\frac{1}{2}\) × 60 × (105 + 50) cm 2
= 30 × 155 cm 2 = 4650 cm 2
For 100 cm 2 cost of glass used = ₹15
∴ For 4650 cm 2 cost of glass = ₹ \(\frac{4650}{100}\) × 15 = ₹ 697.50
Cost of the glass used = ₹ 697.50
Objective Type Questions

Answer:

Given the parallel sides a = 105 cm ; b = 50 cm ; Height = 60 cm
Area of the trapezium = \(\frac{1}{2}\) × h × (a + b)sq. units = \(\frac{1}{2}\) × 60 × (105 + 50) cm 2
= 30 × 155 cm 2 = 4650 cm 2
For 100 cm 2 cost of glass used = ₹15
∴ For 4650 cm 2 cost of glass = ₹ \(\frac{4650}{100}\) × 15 = ₹ 697.50
Cost of the glass used = ₹ 697.50
Objective Type Questions

Q.8The area of the trapezium, if the parallel sides are measuring 8 cm and 10 cm and the height 5 cm is (i) 45 sq. cm (ii) 40 sq. cm (iii) 18 sq. cm (iv) 50 sq. cmv

Solution

(i) 45 sq. cm
Hint: \(\frac{1}{2}\) × h × (a + b) = \(\frac{1}{2}\) × 5 × (10 + 8) = 45
SamacheerKalvi.Guru

Answer:

(i) 45 sq. cm
Hint: \(\frac{1}{2}\) × h × (a + b) = \(\frac{1}{2}\) × 5 × (10 + 8) = 45
SamacheerKalvi.Guru

Q.9In a trapezium if the sum of the parallel sides is 10 m and the area is 140 sq.m, then the height is (i) 7 cm (ii) 40 cm (iii) 14 cm (iv) 28 cmv

Solution

(iv) 28 cm
Hint: Area = \(\frac{1}{2}\) × h × (a + b) = 140 = \(\frac{1}{2}\) × h × 10 ⇒ h = 28

Answer:

(iv) 28 cm
Hint: Area = \(\frac{1}{2}\) × h × (a + b) = 140 = \(\frac{1}{2}\) × h × 10 ⇒ h = 28

Q.10When the non-parallel sides of a trapezium are equal then it is known as (i) a square (ii) a rectangle (iii) an isoceles trapezium (iv) a parallelogramv

Solution

(iii) an isoceles trapezium
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Answer:

Q.1The base of the parallelogram is 16 cm and the height is 7 cm less than its base. Find the area of the parallelogram.v

Solution

In a parallelogram
Given base b = 16 cm; height h = base – 7 cm = 16 – 7 = 9 cm
Area of the parallelogram = (base × height) sq. units
= 16 × 9 cm 2 = 144 cm 2
Area of the parallelogram = 144 cm 2

Answer:

In a parallelogram
Given base b = 16 cm; height h = base – 7 cm = 16 – 7 = 9 cm
Area of the parallelogram = (base × height) sq. units
= 16 × 9 cm 2 = 144 cm 2
Area of the parallelogram = 144 cm 2

Q.2An agricultural field is in the form of a parallelogram, whose area is 68.75 sq. hm. The distance between the parallel sides is 6.25 cm. Find the length of the base.v

Solution

Height of the parallelogram = 6.25 hm
Area of the parallelogram = 68.75 sq. hm
b × h = 68.75
b × 6.25 = 68.75
b = \(\frac{68.75}{6.25}=\frac{6875}{625}\) = 11 km
Length of the base = 11 km.

Answer:

Height of the parallelogram = 6.25 hm
Area of the parallelogram = 68.75 sq. hm
b × h = 68.75
b × 6.25 = 68.75
b = \(\frac{68.75}{6.25}=\frac{6875}{625}\) = 11 km
Length of the base = 11 km.

Q.3A square and a parallelogram have the same area. If the side of the square is 48m and the height of the parallelogram is 18 m. Find the length of the base of the parallelogram.v

Solution

Given side of the square is 48 m
Area of the square = (side × side) sq. unit = 48 × 48 m 2
Height of the parallelogram = 18 m
Area of the parallelogram = ‘bh’ sq. units = b × 18 m 2
Also area of the parallelogram = Area of the square
b × 18 = 48 × 48
b = \(\frac{{48} \times 48}{18 }\) = 8 × 16 = 128 m
Base of the parallelogram = 128 m
SamacheerKalvi.Guru

Answer:

Given side of the square is 48 m
Area of the square = (side × side) sq. unit = 48 × 48 m 2
Height of the parallelogram = 18 m
Area of the parallelogram = ‘bh’ sq. units = b × 18 m 2
Also area of the parallelogram = Area of the square
b × 18 = 48 × 48
b = \(\frac{{48} \times 48}{18 }\) = 8 × 16 = 128 m
Base of the parallelogram = 128 m
SamacheerKalvi.Guru

Q.4The height of the parallelogram is one fourth of its base. If the area of the parallelogram is 676 sq. cm, find the height and the base.v

Solution

Let the base of the parallelogram be ‘b’ cm
Given height = \(\frac{1}{4}\) × base ; Area of the parallelogram = 676 sq. cm
b × h = 676
b × \(\frac{1}{4}\)b = 676
b × b = 676 × 4
b × b = 13 × 13 × 4 × 4
b = 13 × 4 cm = 52 cm
Height = \(\frac{1}{4}\) × 52 cm = 13 cm
Height = 13 cm, Base 52 cm

Answer:

Let the base of the parallelogram be ‘b’ cm
Given height = \(\frac{1}{4}\) × base ; Area of the parallelogram = 676 sq. cm
b × h = 676
b × \(\frac{1}{4}\)b = 676
b × b = 676 × 4
b × b = 13 × 13 × 4 × 4
b = 13 × 4 cm = 52 cm
Height = \(\frac{1}{4}\) × 52 cm = 13 cm
Height = 13 cm, Base 52 cm

Q.5The area of the rhombus is 576 sq. cm and the length of one of its diagonal is half of the length of the other diagonal then find the length of the diagonal.v

Solution

Let one diagonal of the rhombus = d 2 cm
The other diagonal d 2 = \(\frac{1}{2}\) × d 1 cm
Area of the rhombus = 576 sq. cm
\(\frac{1}{2}\) × (d 1 × d 2 ) = 576
\(\frac{1}{2}\) × (d 1 × \(\frac{1}{2}\) d 1 ) = 576
d 1 × d 1 = 576 × 2 × 2 = 6 × 6 × 4 × 4 × 2 × 2
d 1 × d 1 = 6 × 4 × 2 × 6 × 4 × 2
d 1 = 6 × 4 × 2
d 1 = 48 cm
d 2 = \(\frac{1}{2}\) × 48 = 24 cm
∴ Length of the diagonals d 1 = 48 cm and d 2 = 24 cm.

Answer:

Let one diagonal of the rhombus = d 2 cm
The other diagonal d 2 = \(\frac{1}{2}\) × d 1 cm
Area of the rhombus = 576 sq. cm
\(\frac{1}{2}\) × (d 1 × d 2 ) = 576
\(\frac{1}{2}\) × (d 1 × \(\frac{1}{2}\) d 1 ) = 576
d 1 × d 1 = 576 × 2 × 2 = 6 × 6 × 4 × 4 × 2 × 2
d 1 × d 1 = 6 × 4 × 2 × 6 × 4 × 2
d 1 = 6 × 4 × 2
d 1 = 48 cm
d 2 = \(\frac{1}{2}\) × 48 = 24 cm
∴ Length of the diagonals d 1 = 48 cm and d 2 = 24 cm.

Q.6A ground is in the form of isoceles trapezium with parallel sides measuring 42 m and 36 m long. The distance between the parallel sides is 30 m. Find the cost of levelling it at the rate of ₹ 135 per sq. m.v

Solution

Parallel sides of the trapezium a = 42 m; b = 36 m
Also height h = 30 m
Area of the trapezium = \(\frac{1}{2}\) × h × (a + b) sq. unit
= \(\frac{1}{2}\) × 30 × (42 + 36) m 2
= \(\frac{1}{2}\) × 30 × 78 m 2
Area = 1,170 m 2
Cost of levelling 1 m 2 = ₹ 135
∴ Cost of levelling 1170 m 2 = ₹ 1170 × 135 = ₹ 1,57,950
Cost of levelling the ground = ₹ 1,57,950
Challenge Problems

Answer:

Parallel sides of the trapezium a = 42 m; b = 36 m
Also height h = 30 m
Area of the trapezium = \(\frac{1}{2}\) × h × (a + b) sq. unit
= \(\frac{1}{2}\) × 30 × (42 + 36) m 2
= \(\frac{1}{2}\) × 30 × 78 m 2
Area = 1,170 m 2
Cost of levelling 1 m 2 = ₹ 135
∴ Cost of levelling 1170 m 2 = ₹ 1170 × 135 = ₹ 1,57,950
Cost of levelling the ground = ₹ 1,57,950
Challenge Problems

Q.7In a parallelogram PQRS (See the diagram) PM and PN are the heights corresponding to the sides QR and RS respectively. If the area of the parallelogram is 900 sq. cm and the length of PM and PN are 20 cm and 36 cm respectively, find the length of the sides QR and SR. Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.4 1

v

Solution

Considering QR as base of the parallelogram height h 1 = 20 cm
Area of the parallelogram = 900 cm 2
b 1 × h 1 = 900 ; b 1 × 20 = 900
b 1 = \(\frac{900}{20}\) = 45 cm
Again considering SR as base height = 36 cm ; Area = 900 cm 2
b 2 × h 2 = 900 ; b 2 × 36 = 900
b 2 = \(\frac{900}{36}\)
b 2 = 25 cm
SR = 25 cm; QR = 45 cm ; SR = 25 cm

Answer:

Considering QR as base of the parallelogram height h 1 = 20 cm
Area of the parallelogram = 900 cm 2
b 1 × h 1 = 900 ; b 1 × 20 = 900
b 1 = \(\frac{900}{20}\) = 45 cm
Again considering SR as base height = 36 cm ; Area = 900 cm 2
b 2 × h 2 = 900 ; b 2 × 36 = 900
b 2 = \(\frac{900}{36}\)
b 2 = 25 cm
SR = 25 cm; QR = 45 cm ; SR = 25 cm

Q.8If the base and height of a parallelogram are in the ratio 7:3 and the height is 45 cm, then fixed the area of the parallelogram.v

Solution

Given base; height = 7 : 3
Let base = 7x cm
height = 3x cm
also given height = 45 cm
3x = 45 cm 45 .
x = \(\frac{45}{3}\) = 15
Now base = 7x cm = 7 × 15 cm = 105 cm
Area of the parallelogram = b × h sq. unit
= 105 × 45 = 4725 cm 2
= 4725 cm 2
SamacheerKalvi.Guru

Answer:

Given base; height = 7 : 3
Let base = 7x cm
height = 3x cm
also given height = 45 cm
3x = 45 cm 45 .
x = \(\frac{45}{3}\) = 15
Now base = 7x cm = 7 × 15 cm = 105 cm
Area of the parallelogram = b × h sq. unit
= 105 × 45 = 4725 cm 2
= 4725 cm 2
SamacheerKalvi.Guru

Q.9Find the area of the parallelogram ABCD if AC is 24 cm and BE = DF = 8 cm. Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.4 2

v

Solution

Area of the parallelogram ABCD Area of the triangle =Area of the triangle ABC + Area of the triangle ADC
Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.4 3
Area of the parallelogram ABCD = 96 + 96 = 192 cm 2

Answer:

Area of the parallelogram ABCD Area of the triangle =Area of the triangle ABC + Area of the triangle ADC
Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.4 3
Area of the parallelogram ABCD = 96 + 96 = 192 cm 2

Q.10The area of the parallelogram ABCD is 1470 sq. cm. If AB = 49 cm and AD = 35 cm then, find the height, DF and BE. Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.4 4

v

Solution

Area of the parallelogram = 1470 sq. cm
Considering AB = base = 49 cm
height = DF
Area = base × height
49 × DF = 1470
DF = \(\frac{1470}{49}\)
DF = 30 cm
Now considering AD as base
Base = AD = 35 cm ; height = BE
Base × Height = 1470
35 × BE = 1470 : BE = \(\frac{1470}{35}\)
BE = 42 cm ; DF = 30 cm ; BE = 42 cm

Answer:

Area of the parallelogram = 1470 sq. cm
Considering AB = base = 49 cm
height = DF
Area = base × height
49 × DF = 1470
DF = \(\frac{1470}{49}\)
DF = 30 cm
Now considering AD as base
Base = AD = 35 cm ; height = BE
Base × Height = 1470
35 × BE = 1470 : BE = \(\frac{1470}{35}\)
BE = 42 cm ; DF = 30 cm ; BE = 42 cm

Q.11One of the diagonals of a rhombus is thrice as the other. If the sum of the length of the diagonals is 24 cm, then find the area of the rhombus.v

Solution

Let one of the diagonals of rhombus be ‘d 1 ’ cm and the other be d 2 cm.
Give d 1 = 3 × d 2
Also d 1 + d 2 = 24 cm
⇒ 3d 2 + d 2 = 24
4d 2 = 24
d 2 = \(\frac{24}{4}\)
d 2 = 6 cm
d 1 = 3 × d 2 = 3 × 6
d 1 = 18 cm
∴ Area of the rhombus = \(\frac{1}{2}\) × d 1 × d 2 sq. units
= \(\frac{1}{2}\) × 18 × 6 cm 2 = 54 cm 2
Area of the rhombus = 54 cm 2

Answer:

Let one of the diagonals of rhombus be ‘d 1 ’ cm and the other be d 2 cm.
Give d 1 = 3 × d 2
Also d 1 + d 2 = 24 cm
⇒ 3d 2 + d 2 = 24
4d 2 = 24
d 2 = \(\frac{24}{4}\)
d 2 = 6 cm
d 1 = 3 × d 2 = 3 × 6
d 1 = 18 cm
∴ Area of the rhombus = \(\frac{1}{2}\) × d 1 × d 2 sq. units
= \(\frac{1}{2}\) × 18 × 6 cm 2 = 54 cm 2
Area of the rhombus = 54 cm 2

Q.12A man has to build a rhombus shaped swimming pool. One of the diagonal is 13 m and the other is twice the first one. Then find the area of the swimming pool and also find the cost of cementing the floor at the rate of ₹ 15 per sq. cm.v

Solution

Let the first diagonal d 1 = 13 m
d 2 = 2 × 13 m = 26 m
Area of the rhombus = \(\frac{1}{2}\) × d 1 × d 2 sq. units
= \(\frac{1}{2}\) × 13 × 26 m 2 = 169m 2
Cost of cementing 1 m 2 = ₹ 15
Cost of cementing 169 m 2 = ₹ 169 × 15 = ₹ 2,535
Cost of cementing = ₹ 2,535
SamacheerKalvi.Guru

Answer:

Let the first diagonal d 1 = 13 m
d 2 = 2 × 13 m = 26 m
Area of the rhombus = \(\frac{1}{2}\) × d 1 × d 2 sq. units
= \(\frac{1}{2}\) × 13 × 26 m 2 = 169m 2
Cost of cementing 1 m 2 = ₹ 15
Cost of cementing 169 m 2 = ₹ 169 × 15 = ₹ 2,535
Cost of cementing = ₹ 2,535
SamacheerKalvi.Guru

Q.13Find the height of the parallelogram whose base is four times the height and whose area is 576 sq. cm.v

Solution

Let the height be ‘A’ and base be ‘h’ units
Given b = 4 × h
Area of the parallelogram = 576 sq. cm
b × h = 576
4h × h = 576
h × h = \(\frac{576}{4}\) = 144
h × h = 12 × 12
h = 12 cm
Height = 12 cm; base = 4 × 12 = 48 cm

Answer:

Let the height be ‘A’ and base be ‘h’ units
Given b = 4 × h
Area of the parallelogram = 576 sq. cm
b × h = 576
4h × h = 576
h × h = \(\frac{576}{4}\) = 144
h × h = 12 × 12
h = 12 cm
Height = 12 cm; base = 4 × 12 = 48 cm

Q.14The table top is in the shape of a trapezium with measurements given in the figure. Find the cost of the glass used to cover the table at the rate of ₹ 6 per 10 sq. cm. Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.4 5

v

Solution

Length of the parallel sides a = 200 cm
b = 150 cm
Height h = 50 cm
Area of the trapezium = \(\frac{1}{2}\) × h (a + b) sq. units
= \(\frac{1}{2}\) × 50 (200+ 150) cm 2
= \(\frac{1}{2}\) × 50 × 350 cm 2 = 8750 cm 2
Cost for 10 sq. cm glass = ₹ 68
∴ Cost of 8750 cm 2 glass = \(\frac{8750}{10}\) × 6 = ₹ 5250
Cost of glass used = ₹ 5,250

Answer:

Length of the parallel sides a = 200 cm
b = 150 cm
Height h = 50 cm
Area of the trapezium = \(\frac{1}{2}\) × h (a + b) sq. units
= \(\frac{1}{2}\) × 50 (200+ 150) cm 2
= \(\frac{1}{2}\) × 50 × 350 cm 2 = 8750 cm 2
Cost for 10 sq. cm glass = ₹ 68
∴ Cost of 8750 cm 2 glass = \(\frac{8750}{10}\) × 6 = ₹ 5250
Cost of glass used = ₹ 5,250

Q.15Arivu has a land ABCD with the measurements given in the figure. If a portion ABED is used for cultivation (where E is the midpoint of DC). D Find the cultivated area. Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.4 6

v

Solution

From the figure given ABED is a trapazium with height h = 18 m
One of the parallel side a = 24 m
Since E is the midpoint of D.
Other parallel side b = \(\frac{24}{2}\) = 12 m
Area of the cultivated ADEB = \(\frac{1}{2}\) × h(a + b) m 2 = \(\frac{1}{2}\) × 18 (24 + 12)
= 9 × 36 m 2 = 324 m 2
Area of cultivation = 324 m 2
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Answer:

From the figure given ABED is a trapazium with height h = 18 m
One of the parallel side a = 24 m
Since E is the midpoint of D.
Other parallel side b = \(\frac{24}{2}\) = 12 m
Area of the cultivated ADEB = \(\frac{1}{2}\) × h(a + b) m 2 = \(\frac{1}{2}\) × 18 (24 + 12)
= 9 × 36 m 2 = 324 m 2
Area of cultivation = 324 m 2
About Us
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Disclaimer
Contact Us
Facebook
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Copyright © 2026 Samacheer Kalvi Guru