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## Chapter 13: Surface Areas and Volumes

#### TEXTBOOK’S EXERCISE – 13.2

1. The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder.

Sol. Here, h = 14 cm, curved surface area

= 88 cm2, r = ?

The curved surface area of the cylinder = 2πrh

⇒ 88 = 2 × 22/7 × r × 14

⇒ 88 = 44 × 2 × r ⇒ r = 88/44× 2  = 1

Hence, the base diameter of the cylinder

= 1 × 2 cm = 2 cm

2. 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 is required for the same?

Sol. Here, h = 1 m, r = 140/ 2 cm

= 70 cm = 0.7 m

Total surface area of the cylinder

= 2πr (h + r)

= 2 × 22 7 × 0.7 (1 + 0.7) m2

= 44 × 0.1 × 1.7 m2 = 7.48 m2

Hence, 7.48 m2 of sheet is required

#### Class nine Maths NCERT Sol Surface Area Volume

3. 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 figure). Find its:

(i) inner curved surface area,

(ii) outer curved surface area,

(iii) total surface area.

Sol. Here, h = 77 cm,

Outer radius (R) = 4 4 2. cm = 2.2 cm

Inner radius (r) = 42 cm = 2 cm

(i) Inner curved surface area of the pipe = 2πrh = 2 × 22/7 × 2 × 77 cm2

= 2 × 22 × 22 cm2 = 968 cm2

(ii) Outer curved surface area of the pipe

= 2πRh = 2 × 22 7 × 2.2 × 77 cm2

= 44 × 24.2 cm2

= 1064.80 cm2

(iii) Total surface area of the pipe = inner curved surface area + outer curved surface area + areas of the two base rings.

= 2πrh + 2πRh + 2π (R2 – r2)

= 968 cm2 +1064.80 cm2 + 2 × 22/7 [(2.2)2 – 22] cm2

= 2032.80 cm2 + 5.28 cm2 = 2038.08 cm2

4. 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 m2.

Sol. Radius of the roller (r) = 84/2 cm = 42 cm

Length of the roller (h) = 120 cm

Curved surface area of the roller

= 2πrh = 2 × 22/ 7 × 42 × 120 cm2

= 44 × 720 cm2 = 31680 cm2

∴ Area covered by the roller in 1 revolution

= 31680 cm2

∴ Area covered by the roller in 500 revolutions = 31680 × 500 cm2 = 15840000 cm2

Hence, area of the playground

= 15840000/100x 100 m2 = 1584 m2

5. 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 Rs 12.50 per m2.

Sol. Here, r = 50/2 cm = 25 cm = 0.25 m, h = 3.5 m

Curved surface area of the pillar

= 2πrh = 2 × 22/7 × 0.25 × 3.5 m2 = 5.5 m2

Cost of painting 1 m2 = Rs 12.50

∴ Total cost of painting the curved surface of the pillar = Rs 12.50 × 5.5 = Rs 68.75

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