__Exercise 13.3 Page No: 221__

__Exercise 13.3 Page No: 221__

**1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area (Assume π=22/7)**

**Solution:**

Radius of the base of cone = diameter/ 2 = (10.5/2)cm = 5.25cm

Slant height of cone, say l = 10 cm

CSA of cone is = πrl

= (22/7)×5.25×10 = 165

Therefore, the curved surface area of the cone is 165 cm^{2}.

**2. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m. (Assume π = 22/7)**

**Solution:**

Radius of cone, r = 24/2 m = 12m

Slant height, l = 21 m

Formula: Total Surface area of the cone = πr(l+r)

Total Surface area of the cone = (22/7)×12×(21+12) m^{2}

= 1244.57m^{2}

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

**(i) radius of the base and (ii) total surface area of the cone.**

**(Assume π = 22/7)**

**Solution:**

Slant height of cone, l = 14 cm

Let the radius of the cone be r.

(i) We know, CSA of cone = πrl

Given: Curved surface area of a cone is 308 cm^{2}

(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 cone = CSA of cone + Area of base (πr^{2})

Total surface area of cone = 308+(22/7)×7^{2} = 308+154

Therefore, the total surface area of the cone is 462 cm^{2}.

**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.**

**(Assume π=22/7)**

**Solution:**

Let ABC be a conical tent

Height of conical tent, h = 10 m

Radius of conical tent, r = 24m

Let the slant height of the tent be l.

(i) In right triangle ABO, we have

AB^{2 }= AO^{2}+BO^{2}(using Pythagoras theorem)

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

= (10)^{2}+(24)^{2}

= 676

l = 26

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

(ii) CSA of tent = πrl

= (22/7)×24×26 m^{2}

Cost of 1 m^{2} canvas = Rs 70

Cost of (13728/7)m^{2} canvas is equal to Rs (13728/7)×70 = Rs 137280

Therefore, the cost of the canvas required to make such a tent is Rs 137280.

**5. What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6m? 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)

or l = 10

Again, CSA of conical tent = πrl

= (3.14×6×10) m^{2}

= 188.4m^{2}

Let the length of tarpaulin sheet required be L

As 20 cm will be wasted, therefore,

Effective length will be (L-0.2m).

Breadth of tarpaulin = 3m (given)

Area of sheet = CSA of tent

[(L–0.2)×3] = 188.4

L-0.2 = 62.8

L = 63

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

**6. The slant height and base diameter of conical tomb are 25m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs. 210 per 100 m ^{2}. (Assume π = 22/7)**

**Solution:**

Slant height of conical tomb, l = 25m

Base radius, r = diameter/2 = 14/2 m = 7m

CSA of conical tomb = πrl

= (22/7)×7×25 = 550

CSA of conical tomb= 550m^{2}

Cost of white-washing 550 m^{2} area, which is Rs (210×550)/100

= Rs. 1155

Therefore, cost will be Rs. 1155 while white-washing tomb.

**7. A joker’s cap is in the form of right circular cone of base radius 7 cm and height 24cm. Find the area of the sheet required to make 10 such caps. (Assume π =22/7)**

**Solution:**

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)

Or l = 25 cm

CSA of 1 conical cap = πrl

= (22/7)×7×24

= 550

CSA of 10 caps = (10×550) cm^{2} = 5500 cm^{2}

Therefore, the area of the sheet required to make 10 such caps is 5500 cm^{2}.

**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 m ^{2}, what will be the cost of painting all these cones? (Use π = 3.14 and take √(1.04) =1.02)**

**Solution**:

Given:

Radius of cone, r = diameter/2 = 40/2 cm = 20cm = 0.2 m

Height of cone, h = 1m

Slant height of cone is l, and l^{2 }= (r^{2}+h^{2})

Using given values, l^{2} = (0.2^{2}+1^{2})

= (1.04)

Or l = 1.02

Slant height of the cone is 1.02 m

Now,

CSA of each cone = πrl

= (3.14×0.2×1.02)

= 0.64056

CSA of 50 such cones = (50×0.64056) = 32.028

CSA of 50 such cones = 32.028 m^{2}

Again,

Cost of painting 1 m^{2} area = Rs 12 (given)

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

= Rs.384.336

= Rs.384.34 (approximately)

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