Class 9 - Mathematics
Surface Areas and Volumes - Exercise 13.3
Top Block 1
Exercise 13.3
Question :1 : Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. find its curved surface area.
Answer :
Here, r = 10.5/2 cm = 5.25 cm, l = 10 cm.
Curved surface area of the cone = πrl
= (22/7) * 5.25 * 10
= 22 * 7.5 * 10
= 165 cm2
Question : 2: Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Answer :
Here, l = 21 m, r = 24/2 = 12 m
Total surface area of the cone = πr(l + r)
= (22/7) * 12 (21 + 12)
= (22/7) * 12 * 33
= 8712/7
= 1244.57 m2
Question : 3: Curved surface area of a cone is 308 cm2 and its slant height is 14 cm.
Find (i) radius of the base and (ii) total surface area of the cone.
Answer :
Here, l = 14 cm, curved surface area = 308 cm2, r = ?
(i) Curved surface area of the cone = πrl
⇒ 308 = (22/7) * r * 14
⇒ 308 = 22 * r * 7
⇒ 308 = 154 * r
⇒ r = 308/154
⇒ r = 2
Hence, base radius of the cone = 7 cm.
(ii) Total surface area of the cone = πr (l + r)
= (22/7) * 7(14 + 7)
= 22 * 21
= 462 cm2
Question : 4: A conical tent is 10 m high and the radius of its base is 24 m. Find
(i) slant height of the tent.
(ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is Rs 70.
Answer :
Here, h = 10 m, r = 24 m
(i) We have, l2 = h2 + r2
= (10)2 + (24)2
= 100 + 576
= 676
⇒ l = √676
⇒ l = 26 m
(ii) Curved surface area of the tent = πrl
= (22/7) * 24 * 26
Cost of 1 m2 canvas = Rs 70
So, Cost of (22/7) * 24 * 26 m2 of canvas = Rs 70 * (22/7) * 24 * 26
= Rs 10 * 22 * 24 * 260
= Rs 137280
Question : 5: What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for Stitching margins and wastage in cutting is approximately 29 cm (use π = 3.14)
Answer :
Here h = 6 m, r = 8 m
We have, l = √(r2 + h2)
= √(62 + 82)
= √(36 + 64)
= √100
⇒ l = 10 m
Curved surface area of the tent = πrl
= 3.14 * 6 * 10
So, required length of tarpaulin = (3.14 * 6 * 10)/3 m + 20 cm
= (3.14 * 2 * 10) m + 20 cm
= 62.8 + 0.2 m
= 63 m
Mddle block 1
Question : 6: The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white washing its curved surface at the rate of Rs 210 per 100 m2.
Answer :
Here, l = 25 m, r = 14/2 m = 7 m
Curved surface area of the tomb = πrl
= (22/7) * 7 * 25
= 22 * 25
= 550 m2
Cost of white washing 100 m2 = Rs 210
So, Cost of white washing 550 m2 = Rs (210/100) * 550
= 21 * 55
= Rs 1155
Question : 7: A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.
Answer :
Here, r = 7 cm, h = 24 cm
We have, l = √(h2 + r2)
= √(242 + 72)
= √(576 + 49)
= √625
⇒ l = 25 cm
Total curved surface area of 1 cap = πrl
= (22/7) * 7 * 25
= 22 * 25
= 550 cm2
Area of sheet required to make 10 such caps = 10 * 550 = 5500 cm2
Question : 8: A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m.
If the outer side of each of the cones is to be painted and the cost of painting is Rs 12 per m2, what will be the cost of painting all these cones? (Use π = 3.14 and take √1.04 = 1.02)
Answer :
Here, r = 40/2 cm = 20 cm = 0.20 m, h = 1 m
l = √(h2 + r2)
= √{12 + (0.2)2}
= √(1 + 0.4)
= √1.04
⇒ l = 1.02 m
Curved surface area of 1 cone = πrl
Curved surface area of 50 cones = 50 * 3.14 * 0.2 * 1.02
= 32.028 m2
Cost of painting an area of 1 m2 = Rs 12
So, Cost of painting an area of 32.028 m2 = Rs 12 * 32.028
= Rs 384.34 (approx)