Samacheer Class 7 Maths Measurements Intext Questions Book Back Answers & Solutions (Term 1)

📝 Don't just read — test yourselfFree flashcards + scored self-test · no sign-in

Your Progress - Chapter 20% complete

1Book Back Questions53 questions

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

Answer:

ABCD is a parallelogram. It can be divided into two triangles of equal area by drawing the diagonal BD.Area of the parallelogram ABCD = base × height
= AB × DE

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

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.3Count 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). v

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

Answer:

Converting the given parallelograms into rectangles we get.(a) 10 sq. units
(b) 18 sq. units
(c) 16 sq. units
(d) 5 sq. units

Q.4Find the area o the following parallelograms by measuring their base and height, using formula. (e) _____ sq. unitsv

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

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

Q.5Draw as many parallelograms as possible in a grid sheet with the area 20 square units each.v

Answer:

Area of parallelogram (a), (b) or (c) = 20 sq. units
Exercise 2.2
Rhombus
(Try These Textbook Page No. 41)

Q.6Can you find the perimeter of the rhombus?v

Answer:

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

Q.7Can diagonals of a rhombus be of the same length?v

Answer:

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

Q.8A square is a rhombus but a rhombus is not a square. Why?v

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.9Can you draw a rhombus in such a way that the side is equal to the diagonal.v

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.10Can you find the perimeter of the trapezium? Discuss.v

Answer:

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

Q.11Mention any three life situations where the isosceles trapeziums are used?v

Answer:

(i) Glass of a car windows.
(ii) Eye glass (glass in spectacles)
(iii) Some bridge supports.
(iv) Sides of handbags.
About Us
Privacy Policy
Disclaimer
Contact Us

Q.12Find the area and perimeter of the following parallelograms. v

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.13Suresh 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? v

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

Q.14Janaki 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

Answer:

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.15A 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

Answer:

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.16The 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

Answer:

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

Q.17The 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

Answer:

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

Q.18The 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

Answer:

(iii) 13 cm
Hint:

Q.19What 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

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.

Q.20In 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

Answer:

(ii) 192 sq. cm
Hint: Given b = 3 × h; h = 8 cm
Area = b × h = 3h × 8 = 3 × 8 × 8 = 192 cm 2
About Us
Privacy Policy
Disclaimer
Contact Us

Q.21Find the area of rhombus PQRS shown in the following figures. v

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.22Find the area of a rhombus whose base is 14 cm and height is 9 cm.v

Answer:

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.23The 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

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

Q.24A 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

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.25The 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

Answer:

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

Q.26The 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

Answer:

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

Q.27The 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

Answer:

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

Q.28The 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

Answer:

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

Q.29The angle between the diagonals of a rhombus is (i) 120° (ii) 180° (iii) 90° (iv) 100°v

Answer:

(iii) 90°
Hint:
Angles of a rhombus bisect at right angles.
About Us
Privacy Policy
Disclaimer
Contact Us

Q.30Find the area of a trapezium whose parallel sides are 24 cm and 20 cm and the distance between them is 15 cm.v

Answer:

Given the parallel sides a = 24 cm; b = 20 cmDistance 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.31The 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

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∴ The other parallel side = 38 cm.

Q.32The 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

Answer:

Length of the parallel sides a = 55.6 cm ; b = 34.4 cm
Area of the trapezium = 1080 sq. cmDistance between parallel sides = 24 cm.

Q.33The 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

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. cm2a + 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.34The 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

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.35A 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

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.36The 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

Answer:

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

Q.37In 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

Answer:

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

Q.38When 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

Answer:

(iii) an isoceles trapezium
About Us
Privacy Policy
Disclaimer
Contact Us

Q.39The base of the parallelogram is 16 cm and the height is 7 cm less than its base. Find the area of the parallelogram.v

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.40An 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

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.41A 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

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

Q.42The 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

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.43The 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

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.44A 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

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.45In 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. v

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.46If 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

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

Q.47Find the area of the parallelogram ABCD if AC is 24 cm and BE = DF = 8 cm. v

Answer:

Area of the parallelogram ABCD Area of the triangle =Area of the triangle ABC + Area of the triangle ADCArea of the parallelogram ABCD = 96 + 96 = 192 cm 2

Q.48The area of the parallelogram ABCD is 1470 sq. cm. If AB = 49 cm and AD = 35 cm then, find the height, DF and BE. v

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.49One 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

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.50A 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

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

Q.51Find the height of the parallelogram whose base is four times the height and whose area is 576 sq. cm.v

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

Back: Number System Intext Questions

Next: Ch.3