RS Aggarwal Solutions Class 9 Chapter 13 Volume and Surface Area

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RS Aggarwal Solutions Class 9 Chapter 13 Volume and Surface Area

Exercise 13A

Question 1:
(i) length =12cm, breadth = 8 cm and height = 4.5 cm
∴ Volume of cuboid = l x b x h
= (12 x 8 x 4.5) cm³= 432 cm³
∴ Lateral surface area of a cuboid = 2(l + b) x h
= [2(12 + 8) x 4.5] cm²
= (2 x 20 x 4.5) cm² = 180 cm²
∴ Total surface area cuboid = 2(lb +b h+ l h)
= 2(12 x 8 + 8 x 4.5 + 12 x 4.5) cm²
= 2(96 +36 +54) cm²
= (2 x186) cm²
= 372 cm²

(ii) Length 26 m, breadth =14 m and height =6.5 m
∴ Volume of a cuboid = l x b x h
= (26 x 14 x 6.5) m³
= 2366 m³
∴ Lateral surface area of a cuboid =2 (l + b) x h
= [2(26+14) x 6.5] m²
= (2 x 40 x 6.5) m²
= 520 m²
∴ Total surface area = 2(lb+ bh + lh)
= 2(26 x 14+14 x6.5 +26 x6.5)
= 2 (364+91+169) m²
= (2 x 624) m2= 1248 m².

(iii) Length = 15 m, breadth = 6m and height = 5 dm = 0.5 m
∴ Volume of a cuboid = l x b x h
= (15 x 6 x 0.5) m3=45 m³.
∴ Lateral surface area = 2(l + b) x h
= [2(15 + 6) x 0.5] m²
= (2 x 21×0.5) m2=21 m²
∴ Total surface area =2(lb+ bh + lh)
= 2(15 x 6 +6 x 0.5+ 15 x 0.5) m²
= 2(90+3+7.5) m²
= (2 x 100.5) m²
=201 m²

(iv) Length = 24 m, breadth = 25 cm =0.25 m, height = 6m.
∴ Volume of cuboid = l x b x h
= (24 x 0.25 x 6) m³.
= 36 m³.
∴ Lateral surface area = 2(l + b) x h
= [2(24 +0.25) x 6] m²
= (2 x 24.25 x 6) m²
= 291 m².
∴ Total surface area =2(lb+ bh + lh)
=2(24 x 0.25+0.25x 6 +24 x 6) m²
= 2(6+1.5+144) m²
= (2 x151.5) m²
=303 m².

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Question 2:
Length of Cistern = 8 m
Breadth of Cistern = 6 m
And Height (depth) of Cistern =2.5 m
∴ Capacity of the Cistern = Volume of cistern
∴ Volume of Cistern = (l x b x h)
= (8 x 6 x2.5) m³
=120 m³
Area of the iron sheet required = Total surface area of the cistem.
∴ Total surface area = 2(lb +bh +lh)
= 2(8 x 6 + 6×2.5+ 2.5×8) m²
= 2(48 + 15 + 20) m²
= (2 x 83) m2=166 m²

Question 3:
Length of a room =9m,
Breadth of a room = 8m
And height of room = 6.5 m
∴ Area of 4 walls = Lateral surface area
= 2 (l+ b) x h
= [2 (9+8) x 6.5] m²
= (2 x 17 x 6.5) m²
=221 m²
∴ Area not be whitewashed = (area of 1 door) + (area of 2 windows)
= (2 x 1.5) m² + (2 x 1.5 x 1) m²
= 3m²+ 3m²=6m²
∴ Area to be whitewashed = (221-6) m² =215 m²
∴ Cost of whitewashing the walls at the rate of Rs.6.40 per
Square meter = Rs. (6.40 x 215) = Rs. 1376

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Exercise 13B

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Exercise 13C

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Exercise 13D

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Complete RS Aggarwal Solutions Class 9

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