**Selina Concise Class 7 Math Chapter 20 Mensuration Exercise 20A Solution**

**EXERCISE 20A**

(1) Perimeter = 2 × (135 + 65)m

= 2 × 200 m = 400m

Rate of fencing = Rs 60/m

∴Total cost of fencing = Rs (400 × 60) = Rs 24000

(2) Let the length will be 7x m and breadth will be 4x m.

Then, 2 × (7x + 4x) = 440

⇒ 2 × 11x = 440

⇒ 22x = 440

⇒ x = 440/22

⇒ x = 20

Therefore, length = (7 × 20) = 140 m and breadth = (4 × 20) = 80m

Rate of fencing = Rs 150/m

∴ Total cost of fencing = Rs (150 × 440) = Rs 66,000

(3) Let the breadth will be x m.

Applying Pythagoras theorem,

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

⇒ (34)^{2} = (30)^{2} + x^{2}

⇒ x^{2} = (34)^{2 }– (30)^{2}

⇒ x^{2} = 1156 – 900

⇒ x^{2} = 256

⇒ x = √256 = 16

Therefore, breadth = 16 m.

∴ Perimeter = 2 × (30 + 16) m

= 2 × 46 m = 92 m

(4) Let the each side will be x m.

Applying Pythagoras theorem,

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

⇒ (12√2)^{2} = x^{2} + x^{2}

⇒ (144 × 2) = 2x^{2}

⇒ 2x^{2} = 288

⇒ x^{2} = 288/2

⇒ x = √144 = 12

Therefore, length of each side = 12 m

∴ Perimeter = (4 × 12) m = 48 m.

(5) Here, 16 dm = 1.6 m

∴ Perimeter = 2 × (22.5 + 1.6) m

= 2 × 24.1 m = 48.2 m

(6) Let the breadth will be x cm.

Applying Pythagoras theorem,

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

⇒ (25)^{2} = (24)^{2} + x^{2}

⇒ x^{2} = (25)^{2} – (24)^{2}

⇒ x^{2} = 625 – 576

⇒ x = √49 = 7

Breadth = 7 cm

∴ Perimeter = 2 × (24 + 7) cm

= 2 × 31 cm = 62 cm

(7) Let the length and the breadth will be 5x and 3x m respectively.

Perimeter = 2 × (5x + 3x) m = (2 × 8x) m = 16x m

Rate of fencing = Rs 48/m

Total cost of fencing = Rs (16x × 48) = Rs 768x

But given = Rs 19200

∴ 768x = 19200

⇒ x = 19200/768

⇒ x = 25

Therefore, Length = (5 × 25) = 125m

And breadth = (3 × 25) = 75 m

(8) Perimeter of square shape = (4 × 20) cm = 80 cm

Let the breadth of rectangular shape be x cm.

Then, Perimeter of rectangular shape = 2 × (24 + x) cm

Then, 2 × (24 + x) = 80

⇒24 + x = 80/2

⇒ x = 40 – 24 = 16

Therefore, breadth = 16 cm.

(9) P = perimeter, Length = l and breadth = b

(i) P = 2 × (l + b)

⇒ P = 2 × (38 + 27)

⇒ P = 2 × 65 = 130 cm

(ii) 2 × (l + b) = P

⇒ (24 + b) = 88/2

⇒ b = 44 – 24

⇒ b = 20

(iii) 2 × (l + b) = P

⇒ l + 28 = 96/2

⇒ l = 48 – 28

⇒ l = 20

(10) Let the length of each side will be x m.

Perimeter = 4x m

Rate of fencing = Rs 75/m

Total cost of fencing = Rs (4x × 75)

But given = Rs 67500

∴ 4x × 75 = 67500

⇒ 4x = 67500/75

⇒ 4x = 900

⇒ x = 900/4

⇒ x = 225

Hence, the side of square = 225m.

(11) Perimeter of rectangular field = 2 × (36 + 28) cm

= 2 × 64 cm = 128 cm

Given, its perimeter = perimeter of a square

Let the length of each side will be x cm.

Then, its perimeter = 4x cm

∴ 4x = 128

⇒ x = 128/4

⇒ x = 32

__Get next Exercise 20B solutions click here__

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