Class nine Maths NCERT Sol Surface Area Volume
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