# NCERT Solutions for Class 9th Mathematics

## Chapter 13. Surface Areas and Volumes

### Exercise 13.2

**EXERCISE 13.2**

**Assume that **π **= **** , unless stated otherwise.**

**Q1. The curved surface area of a right circular cylinder of height 14 cm is 88 cm ^{2}. Find the diameter of the base of the cylinder.**

**Solution: **

H = 14cm, curved surface area of cylinder = 88cm^{2}

Curved surface area of cylinder = 88cm^{2}

⇒ 2 × × r × h = 88cm^{2}

⇒ 2 × × r × 14 = 88cm^{2}

⇒ 88r = 88

⇒ r = 88/88

⇒ r = 1cm

D = 2r ⇒ 2 ×1 = 2cm

**Q2. It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square metres of the sheet are required for the same? **

**Solution:**

D = 140cm ⇒ r = 70cm = 0.7m, H = 1m

Total surface area of cylinder = 2πr(r + h)

⇒ 2 × × 0.7 (0.7 + 1)

⇒ 2 × 22 × 0.1 × 1.7

⇒ 7.48m^{2}

Hence, 7.48m^{2} metal sheet required to make closed cylindrical tank.

**Q3. A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm (see Fig. 13.11). Find its**

**(i) Inner curved surface area, **** **** **

**(ii) Outer curved surface area, **

**(iii) Total surface area.**

**Solution:**

H = 77cm, H = 77cm

D = 4cm, D = 4.4cm

R = 2cm, R = 2.2cm

(i) interior curved surface area = 2πrh

⇒ 2 × × 2 ×77

⇒ 88 × 11

⇒ 968 cm^{2}

(ii) exterior curved surface area = 2πrh

⇒ 5.28 cm^{2}

Total surface area = interior surface area + exterior surface area + ring surface area

⇒ 1064.8 + 9.68 + 5.28 cm^{2}

⇒ 2038.08 cm^{2}

**Q4. The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m ^{2}.**

**Solution:**

D = 84cm ⇒ r = 42cm, H = 120cm

Curved surface area of cylinder = 2πrh

⇒ 2 × × 42 × 120

⇒ 44 × 6 × 120

⇒ 31680 cm^{2}

Total area of playground

**Q5. A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of 12.50 per m ^{2}.**

**Solution:**

D = 50 cm , ⇒ R = 0.25m, H = 3.5m

Curved surface area of cylinder = 2πrh

**Q6. Curved surface area of a right circular cylinder is 4.4 m ^{2}. If the radius of the base of the cylinder is 0.7 m, find its height.**

**Solution:**

Curved surface area of cylinder = 4.4 m^{2} , radius = 0.7 m

Let the height of the cylinder be = h

2πrh = 4.4

**Q7. The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find**

**(i) its inner curved surface area,**

**(ii) the cost of plastering this curved surface at the rate of 40 per m ^{2}.**

**Solution:**

Inner diameter of circular well = 3.5 m ⇒ r = 1.75 m

Depth of the well = 10 m

(i) inner surface area of the well = 2πrh

**Q8. In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.**

**Solution:**

The length of the pipe = 28 m

**Q9. Find**

**(i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high.**

**(ii) how much steel was actually used, if ****of the steel actually used was wasted in making the tank.**

**Solution:**

(i) diameter of the cylindrical petrol tank = 4.2 m

Radius of the tank = 2.1m , height = 4.5 m

Curved surface area of cylinder = 2πrh

**Q10. In Fig. 13.12, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade.**

**Solution:**

Height of the folding = 2.5 cm

Height of the frame = 30 cm

Diameter = 20 cm ⇒ radius = 10 cm

Now cloth required for covering for lampshade

= C.S.A of top + C.S.A of middle + C.S.A of bottom

**Q11. The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?**

**Solution:**

Radius of pen holder = 3 cm

Height of pen holder = 10.5 cm

Cardboard required for pen holder = CSA of pen holder + area of circular base

⇒ 2πrh + πr^{2}

⇒ πr(2h + r)