Here you can get solutions of RS Aggarwal Solutions Class 9 Chapter 13 Volume and Surface Area. These Solutions are part of RS Aggarwal Solutions Class 9. we have given RS Aggarwal Solutions Class 9 Chapter 13 Volume and Surface Area download pdf.

## 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) m
=303 m2.

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
= 3m+ 3m=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

Question 4: Question 5: Question 6: Question 7: Question 8: Question 9: Question 10: Question 11: Question 12: Question 13: Question 14: Question 15: Question 16: Question 17: Question 18: Question 19: Question 20: ### Exercise 13B

Question 1: Question 2: Question 3: Question 4: Question 5: Question 6: Question 7: Question 8: Question 9: Question 10: Question 11: Question 12: Question 13: Question 14: Question 15: Question 16: Question 17: Question 18: ### Exercise 13C

Question 1: Question 2: Question 3: Question 5: Question 6: Question 7: Question 8: Question 9: Question 10: Question 11: Question 12: Question 13: Question 14: Question 15: Question 16: Question 17: Question 18: ### Exercise 13D

Question 1: Question 2: Question 3: Question 4: Question 5: Question 6: Question 7: Question 8: Question 9: Question 10: Question 11: Question 12: Question 13: Question 14: Question 15: Question 16: Question 17: Question 18: Question 19: Question 20: Question 21: Question 22: Question 23: Complete RS Aggarwal Solutions Class 9

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RS Aggarwal Solutions Class 9 Chapter 13 Volume and Surface Area Reviewed by Admin on November 15, 2020 Rating: 5