## NCERT solutions for class 9 Maths

## Surface Areas and Volumes

**Chapter 13**

**Exercise 13.3**

**EX 13.2 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:**

Radius (r) = cm = 5.25 cm

and slant height *(l)* =10 cm

Curved surface area of the cone = πr*l*

= x x 10cm^{2}

= 11 x 15 x 1 cm^{2} = 165cm^{2}

**Ex 13.3 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:**

Radius (r) = m = 12 m

and slant height *(l)* = 21 m

∴ Total surface area of a cone = πr(r *+l)*

= * 12 * (12 + 21 ) = = * 12 * 33 = 1244.57 m^{2}

**Ex 13.3 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.**

**Solution:**

Curved surface area = πr*l* = 308 cm^{2}

Slant height *(l)* = 14 cm

(i) Let the radius of the base of cone = r cm

= 308 ⇒ x r x 14 = 308

r = = 7cm

Hence, the radius of the cone is 7 cm

(ii) Total surface area of the cone = πr(r + *l*)

= * 7 * (7 + 14) = 22 * 21 = 462 cm^{2}

**Ex 13.3 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 ₹70.**

**Solution:**

Height tent (h) = 10 m

Radius of the cone(r) = 24 m

(i) The slant height, *l *m

⇒*l ^{2} *= r

^{2}+ h

^{2}⇒

*l*= 24

^{2}^{2}+ 10

^{2}= 576 + 100= 676

⇒

*l*= m = 26m

(ii) Curved surface area of the cone = πrl

= * 24 * 26 m^{2}

Cost of 1m^{2 }canvas =₹70

∴ The cost of * 24 * 26 m^{2} canvas = 70 * * 24 * 26 = ₹137280

**Ex 13.3 Question 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:**

Radius of cone (r) = 6 m

Height(h) = 8m

∴ Slant height (l) = *l ^{2} *= r

^{2}+ h

^{2}

⇒

*l*= 6

^{2}^{2}+ 8

^{2}= 36 + 64 = 100

⇒

*l*= m = 10 m

Area of tarpaulin to make the tent = πr

*l*

= 3.14 x 6 x 10 = 1884 m

^{2}

Let the length of 3 m wide tarpaulin = L m

Therefore, the area of tarpaulin required = 3 * L

According to question,

3 * L = 188.4 ⇒ L = = 62.80 m

Extra tarpaulin for stitching margins and wastage = 20cm = 0.2m

Thus, total length of tarpaulin = 62.8 m + 0.2 m = 63 m

**Ex 13.3 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 ₹ 210 per 100 m ^{2} .**

**Solution:**

Radius of conical tomb (r) = m = 7 m

Slant height

*(l)*= 25 m

∴ Curved surface area of conical tomb = πr

*l*

= x 7 x 25 m

^{2}= 550 m

^{2}

Cost of white-washing for 100 m

^{2}area = Rs. 210

∴ Cost of white-washing for 550 m

^{2}area

= Rs. x 550 = Rs. 1155

**Ex 13.3 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:**

Radius of right circular cone (r) = 7 cm and height (h) = 24 cm

Slant height *(l)* = *l* m

⇒*l ^{2} *= r

^{2}+ h

^{2}=

*l*= 7

^{2}^{2}+ 24

*= 49 + 576 = 625.*

^{2}⇒ * l *= cm = 25 cm

∴Area of sheet required to make 1 cap = πrl = x 7 x 25 cm^{2} = 550 cm^{2}

∴ Area of sheet required to make 10 caps = 10 x 550 cm^{2}

= 5500 cm^{2}

**Ex 13.3 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 ₹12 per m², what will be the cost of painting all these cones? (Use π = 3.14 and take = 1.02)**

**Solution:**

Radius of cone (r) = 40/2 = 20 cm and height (h) = 1 cm

Slant height *(l)* = *l* m

⇒*l ^{2} *= r

^{2}+ h

^{2}=

*l*= (0.2)

^{2}^{2}+ 1

*= 0.04+ 1 = 1.04*

^{2}⇒ * l *= = 1.02m

Curved surface area of cone = πrl = 3.14 * 7 * 25 cm^{2} = 6.4056 m^{2}

∴ Curved surface area of 50 cone s = 50 x 6.4056 = 32.028 m^{2}

Cost of painting at rate of ₹12 per m^{2} = ₹12 * 32.028 = Rs. 384.34 (approx.)