RS Aggarwal Solutions for Class 7 Maths Exercise 20G Chapter 20 Mensuration in simple PDF are given here. This exercise of RS Aggarwal Solutions for Class 7 Chapter 20 contains objective type questions. It covers all the topics in this chapter. Students can learn more about these topics by solving the problems of RS Aggarwal Solutions for Class 7 Maths Chapter 20 Mensuration Exercise 20G with the help of solutions provided here.

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**Mark against the correct answer in each of the following:**

**1. The length of a rectangle is 16 cm and the length of its diagonal is 20 cm. The area of the rectangle is**

**(a) 320 cm ^{2} (b) 160 cm^{2} (c) 192 cm^{2} (d) 156 cm^{2}**

**Solution:-**

(c) 192 cm^{2}

Because,

Let ABCD be the rectangular plot.

Then, AB = 16 cm and AC = 20 cm

BC =?

According to Pythagoras theorem,

From right angle triangle ABC, we have:

= AC^{2} = AB^{2} + BC^{2}

= 20^{2} = 16^{2} + BC^{2}

= BC^{2} = 20^{2} – 16^{2}

= BC^{2} = 400 â€“ 256

= BC^{2} = 144

= BC = âˆš144

= BC = 12 cm

Hence, the area of the rectangle plot = (l Ã— b)

Where, l = 16 cm, b = 12 cm

Then,

= (16 Ã— 12)

= 192 cm^{2}

**2. Each diagonal of a square is 12 cm long. Its area is**

**(a) 144 cm ^{2} (b) 72 cm^{2} (c) 36 cm^{2} (d) none of these**

**Solution:-**

(b) 72 cm^{2}

Because,

Given,

Diagonal of a square = 12 cm

Area of a square = (1/2) Ã— (diagonal)^{2}

= (1/2) Ã— 12 Ã— 12

= 1 Ã— 6 Ã— 12

= 72cm^{2}

**3. The area of a square is 200 cm ^{2}. The length of its diagonal is**

**(a) 10 cm (b) 20 cm (c) 10âˆš2 cm (d) 14.1 cm**

**Solution:-**

(b) 20 cm

Because,

We know that,

Area of a square = (1/2) Ã— (diagonal)^{2}

200 = (1/2) Ã— (diagonal)^{2}

(Diagonal)^{2} = 200 Ã— (2/1)

Diagonal = âˆš400

Diagonal = 20 cm

**4. The area of a square field is 0.5 hectare. The length of its diagonal is**

**(a) 100 m (b) 50 m (c) 250 m (d) 50âˆš2 m**

**Solution:-**

(a) 100 m

Because,

1 hectare = 10000 m^{2}

âˆ´ Area of square field = 0.5 Ã— 10000 = 5000 m^{2}

WKT, Area of a square = (1/2) Ã— (diagonal)^{2}

5000 = (1/2) Ã— (diagonal)^{2}

(Diagonal)^{2} = 5000 Ã— (2/1)

Diagonal = âˆš10000

Diagonal = 100 m

**5. The length of a rectangular field is thrice its breadth and its perimeter is 240 m. The length of the field is**

**(a) 80 m (b) 120 m (c) 90 m (d) none of these**

**Solution:-**

(c) 90 m

Because,

Let us assume the Breadth of the rectangular field be x m.

Let us assume the length of the rectangular field is thrice its breadth be 3x m

Then,

Perimeter of the rectangular field = 2(l + b)

= 240 = 2(x + 3x)

= 240 = 2(4x)

= 240 = 8x

= x = 240/8

= x = 30

âˆ´the length of the rectangular field is 3x = 3 Ã— 30 = 90 m

**6. On increasing each side of a square by 25 %, the increase in area will be**

**(a) 25% (b) 55% (c) 40.5% (d) 56.25%**

**Solution:-**

(d) 56.25%

Because,

Let the length of each side be a cm

Then, its area = a^{2} cm^{2}

Increased side = (a + 25% 0of a) cm

= (a + (1/4) a)

= (5/4)a cm

Area of the square = ((5/4)a)^{2}

= (25/16)a^{2 }cm^{2}

Increase in area = [(25/16)a^{2} â€“ a^{2}] cm

= (25a^{2} â€“ 16a^{2})/ (16)

= (9a^{2}/16) cm^{2}

% increase in the area = (Increased area / old area) Ã— 100

= [(9a^{2}/16)/ a^{2}] Ã— 100

= (9 Ã— 100)/ 16

= 56.25%

**7. The area of a square and that of a square drawn on its diagonal are in the ratio**

**(a) 1: âˆš2 (b) 1: 2 (c) 1: 3 (d) 1: 4**

**Solution:-**

(b) 1: 2

Because,

Let the side of square be a.

And Length of its diagonal = âˆš2 a

âˆ´Required ratio = a^{2}/ (âˆš2a)^{2}

= a^{2}/2a^{2}

= Â½

= 1: 2

**8. The perimeters of a square and a rectangle are equal. If their areas be A m ^{2} and B m^{2}, then which of the following is a true statement?**

**(a) A < B (b) A < B (c) A > B (d) A > B**

**Solution:-**

(c) A > B

Because,

(area of the square) > (area of the rectangle), when perimeters are equal.

**9. The length and breadth of a rectangular field are in the ratio 5: 3 and its perimeter is 480 m. The area of the field is**

**(a) 7200 m ^{2} (b) 13500 m^{2} (c) 15000 m^{2} (d) 540000 m^{2}**

**Solution:-**

(b) 13500 m^{2}

Because,

Let the length of the rectangular field = 5x

The breadth of the rectangular field = 3x

Then,

perimeter of the rectangular land = 2 ( l + b )

480= 2 (5x + 3x)

480 = 2 (8x)

480 = 16x

x = (480/16)

x = 30

âˆ´ the length of the rectangular field = 5x = 5 Ã— 30 = 150m

The breadth of the rectangular field = 3x = 3 Ã— 30 = 90m

Area of rectangular field = (l Ã— b) sq units

Then,

= (150 Ã— 90)

= 13500 m^{2}

**10. The length of a room is 15 m. the cost of carpeting it with a carpet 75 cm wide at â‚¹ 50 per meter is â‚¹ 6000. The width of the room is**

**(a) 6 m (b) 8 m (c) 13.4 m (d) 18 m**

**Solution:-**

(a) 6 m

Because,

From the question,

Total cost of carpeting = â‚¹ 6000

Rate of carpeting = â‚¹ 50 per meter

Then,

Length of the carpet = (6000/50) = 120 m

âˆ´Area of the carpet = (120 Ã— (75/100))

= 90 m^{2}

Area of the floor = Area of the carpet = 90 m^{2}

Hence, the width of the room = (Area/length)

= (90/15)

= 6 m

**11. The sides of a triangle measures 13 cm, 14 cm and 15 cm. Its area is**

**(a) 84 cm ^{2} (b) 91 cm^{2} (c) 168 cm^{2} (d) 182 cm^{2}**

**Solution:-**

(a) 84 cm^{2}

Because,

Let a, b, c be the sides of the triangle

Then, s = (1/2) (a + b + c)

= (1/2) Ã— (13 + 14 + 15)

= (42/2)

= 21

Area of triangle = âˆš(s(s-a) Ã— (s-b) Ã— (s-c))

= âˆš(21 Ã— (21 – 13) Ã— (21 – 14) Ã— (21 – 15))

= âˆš(21 Ã— (8) Ã— (7) Ã— (6))

= âˆš(3 Ã— 7 Ã— 2 Ã— 2 Ã— 2 Ã— 7 Ã— 2 Ã— 3)

= âˆš(3^{2} Ã— 7^{2} Ã— 2^{2 }Ã— 2^{2})

By cancelling square and square root we get,

= (3 Ã— 7 Ã— 2 Ã— 2)

= 84 cm^{2}

**12. The base and height of a triangle is are 12 m and 8 m respectively. Its area is**

**(a) 96 m ^{2} (b) 48 m^{2} (c) 16âˆš3 m^{2} (d) 16âˆš2 m^{2}**

**Solution:-**

(b) 48 m^{2}

Because,

Given,

base = 12 m

height = 8 m

âˆ´ Area of the triangle = (1/2) Ã— base Ã— height

= (1/2) Ã— 12 Ã— 8

= 1 Ã— 6 Ã— 8

= 48 m^{2}

**13. The area of an equilateral triangle is 4âˆš3 cm ^{2}. The length of each of its sides is**

**(a) 3 cm (b) 4 cm (b) 2âˆš3 cm (d) Â½ âˆš3 cm**

**Solution:-**

(b) 4 cm

Because,

From the question is given that,

Area of an equilateral triangle is (4âˆš3) cm^{2}

Area of the equilateral triangle = ((âˆš3)/4) Ã— (sides)^{2} sq. Units

= (4âˆš3) = ((âˆš3)/4) Ã— (sides)^{2}

= (sides)^{2} = (4âˆš3)/ (4/(âˆš3)

= (sides)^{2 }= (4âˆš3 Ã— 4)/ âˆš3

= (sides)^{2 }= 4 Ã— 4

= Sides = âˆš16

= Sides = 4 cm

Hence, the length of each side of the triangle is 4 cm.

**14. Each side of an equilateral triangle is 8 cm long. Its area is**

**(a) 32 cm ^{2} (b) 64 cm^{2} (c) 16âˆš3 cm^{2} (d) 16âˆš2 cm^{2}**

**Solution:-**

(c) 16âˆš3 cm^{2}

Because,

Area of the equilateral triangle = ((âˆš3)/4) Ã— (sides)^{2} sq. Units

= (âˆš3 /4) Ã— (8)^{2}

= (âˆš3/4) Ã— 64)

= (âˆš3 Ã— 16)

= 16âˆš3 cm^{2}

**15. The height of an equilateral triangle is âˆš6 cm. Its area is**

**(a) 3âˆš3 cm ^{2} (b) 2âˆš3 cm^{2} (c) 2âˆš2 cm^{2} (d) 6âˆš2 cm^{2}**

**Solution:-**

(b) 2âˆš3 cm^{2}

Because,

Let an equilateral triangle with one side of length a cm

Diagonal of an equilateral triangle = (âˆš3/2) a

= (âˆš3/2) a = âˆš6

= a = (âˆš6 Ã— 2)/ âˆš3

= a = (âˆš3 Ã— âˆš2 Ã— 2)/ âˆš3

= a = 2âˆš2

Area of equilateral triangle = âˆš3/4 a^{2}

^{ } = âˆš3/4 Ã— (2âˆš2)^{2}

^{ } = (âˆš3/4) Ã— 8

^{ } = 2âˆš3 cm^{2}

**16. One side of a parallelogram is 16 cm and the distance of this side from the opposite side is 4.5 cm. The area of the parallelogram is**

**(a) 36 cm ^{2} (b) 72 cm^{2} (c) 18 cm^{2} (d) 54 cm^{2}**

**Solution:-**

(b) 72 cm^{2}

Because,

From the question is given that,

Base of the parallelogram = 16 cm

Height of the parallelogram = 4.5 cm

âˆ´area of the parallelogram = base Ã— height

= 16 Ã— 4.5

= 72 cm^{2}