# Class 9 NCERT Solutions – Chapter 13 Surface Areas And Volumes – 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.**

**Solution:**

Given:

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free classeswhich will definitely help them in making a wise career choice in the future.Radius = diameter / 2

Radius(r) = (10.5/2)cm = 5.25cm

Slant height of cone, (l) = 10 cm

Since, Curved surface area of cone is = πrl

= (22/7)×5.25×10 = 165 cm

^{2}Therefore, the curved surface area of the cone of height 10m and base 10.5 cm is 165 cm

^{2}.

**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.**

**Solution:**

Given:

Radius = diameter / 2

Radius (r) = 24/2 m = 12m

Slant height, (l) = 21 m

Since, Total surface area of the cone = πr(l+r)

= (22/7)×12×(21+12) m

^{2}= 1244.57m

^{2}Therefore, the total surface area of the cone with height 21m and base 24m is 1244.57m

^{2}.

**Question 3: Curved surface area of a cone is 308 cm**^{2} and its slant height is 14 cm. Find

^{2}and its slant height is 14 cm. Find

**(i) radius of the base and **

**(ii) total surface area of the cone.**

**Solution:**

Given:

Slant height (l) = 14 cm

Curved surface area = 308 cm

^{2}radius = r.

(i) radius of the base

Curved surface area = 308 cm

^{2}Curved surface area of cone = πrl

(308) = (22/7)×r×14

308 = 44 r

r = 308/44 = 7

Radius of a cone base is 7 cm.

(ii) total surface area of the cone.

Total surface area of cone = πr(l+r)

Total surface area of cone = 308+(22/7)×72

= 308+154

= 462 cm

^{2}Therefore, the total surface area of the cone with CSA as 308 cm

^{2}is 462 cm^{2}.

**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 m^{2} canvas is Rs. 70.

**Solution:**

Given:

Height of conical tent (h) = 10m

Radius of conical base (r) = 24m

Slant height of the tent = l.

(i) slant height of the tent.

l

^{2}= h^{2 }+ r^{2 }[using Pythagoras theorem]= (10)

^{2 }+ (24)^{2}= 676

l = 26

Therefore, the slant height of the tent = 26 m.

(ii) cost of the canvas required to make the tent, if the cost of 1 m

^{2}canvas is Rs. 70.Curved surface area of tent = πrl

= (22/7) × 24 × 26 m

^{2}As cost of 1 m

^{2}canvas = Rs 70Cost of (13728/7)m

^{2}canvas = (13728/7)×70= Rs 137280

Therefore, the cost of the canvas required to make the tent is 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 20 cm (Use** π** = 3.14).**

**Solution:**

Height of conical tent (h) = 8m

Radius of base of tent (r) = 6m

Slant height of tent, l

^{2}= r^{2}+h^{2}l

^{2}= 6^{2}+8^{2}= 36+64

= 100 [Taking square root on both the side]

l = 10

As, Curved surface area = πrl

= (3.14×6×10) m

^{2}= 188.4m

^{2}Let the length of tarpaulin sheet required = L.

As 20 cm will be wasted,

Therefore, the Effective length will be (L – 0.2m).

Given breadth of tarpaulin = 3m

Area of sheet = Curved surface area of tent

[(L – 0.2) × 3] = 188.4

L – 0.2 = 62.8

L = 63 m

Therefore, the length of the required tarpaulin sheet will be 63 m.

**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 m**^{2}.

^{2}.

**Solution:**

Given:

Slant height of conical tomb (l) = 25m

Base radius (r) = diameter/2

= 14/2 m

= 7m

Curved surface area of conical tomb = πrl

= (22/7)×7×25

= 550

Curved surface area of conical tomb= 550m

^{2}Cost of white-washing 100 m

^{2}area = Rs 210Cost of white-washing 550 m

^{2}area = Rs (210×550)/100= Rs. 1155

Therefore, the cost will be Rs. 1155 while white-washing tomb of area 550 m

^{2}.

**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.**

**Solution:**

Given:

Radius of conical cap (r) = 7 cm

Height of conical cap (h) = 24cm

Slant height, l

^{2}= r^{2}+h^{2}= 7

^{2}+24^{2}= 49+576

= 625 [Taking square root on both the side]

l = 25 cm

Curved surface area of 1 conical cap = πrl

= (22/7)×7×24

= 550

Curved surface area of 10 caps = (10×550) cm

^{2}= 5500 cm^{2}Therefore, the area of 5500 cm

^{2}of the sheet is required to make 10 such caps.

**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 take1.04 = 1.02) **

**Solution:**

Given:

Radius of cone (r) = diameter/2

= 40/2 cm

= 20cm

= 0.2 m

Height of cone (h) = 1m

Slant height, l

^{2}= r^{2 }+ h^{2}Using given values, l

^{2}= (0.2^{2}+1^{2})= (1.04)

l = 1.02

Slant height of the cone (l) = 1.02 m

Now,

Curved surface area of each cone = πrl

= 3.14×0.2×1.02

= 0.64056

Curved surface area of 50 such cones = (50×0.64056) = 32.028

Curved surface area of 50 such cones = 32.028 m

^{2}Also given Cost of painting 1 m

^{2}area = Rs 12Cost of painting 32.028 m

^{2}area = Rs (32.028×12)= Rs. 384.336

= Rs.384.34 (approximately)

Therefore, the cost of painting all the cones is Rs. 384.34.