CBSE Class 9 Maths - Surface Areas and Volumes - NCERT Exercise 13.2 (#cbseclass9Notes)

Surface Areas and Volumes

NCERT Exercise 13.2

Q1: 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.

Answer: Given, h = 14 cm,
curved, surface area (CSA) = 88 cm², r = ?
CSA  = 2πrh
⇒ 88 = 2 × (22/7) × r × 14
⇒ 88 = 44 × 2 × r
⇒ r = 88/88 = 1
∴ diameter = 2r = 2 cm


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?.



Answer: Given, h = 1m
r = 140 / 2 = 70cm = 0.7m
TSA (cylinder) = 2πr (h + r)
= 2 × (22/7) × 0.7(1 + 0.7)
= 44 × 0.1 × 1.7
= 7.48 m²  (Answer)


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 figure). Find its.
(i) inner curved surface area,
(ii) outer curved surface area,
(iii) total surface area.


Answer: Given, h = 77 cm,
Outer radius (R) = 4.4/2 = 2.2 cm,
Inner radius (r) = 4/2 = 2 cm

(i) Inner curved surface area (CSA₁)) of the pipe
= 2πrh
= 2 × (22/7) × 2 × 77
= 2 × 22 × 22 = 968 cm² (Answer)

(ii) Outer curved surface area (CSA₂) of the pipe = 2πRh
= 2 ×(22/7) × 2.2 × 77
= 44 × 24.2
= 1064.80 cm² (Answer)

(iii) Total surface area of the pipe = CSA₁ + CSA₂ + areas of the two base rings.

= 968 + 1064.80 + 2π (R² – r²)

= 2032.80 + 2(22/7)(2.2² - 2²)

= 2032.80 + 5.28 

= 2038.08 cm² (Answer)

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 m².

Answer: Radius of roller (r) = 84/2 = 42 cm
Length of the roller (h) = 120 cm
CSA = 2πrh
= 2 × (22/7) × 42 × 120
= 44 × 720
= 31680 cm²
∴ area covered by the roller in 1 revolution = 31680 cm²
area covered by the roller in 500 revolutions = 31680 × 500 = 15840000 cm²

Hence, area of of the playground = 15840000 / 10,000 = 1584 m² (Answer)


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 Rs 12.50 per m².

Answer: Given,
r = 50/2 = 25 cm = 0.25 m, h = 3.5 m
CSA of pillar = 2πrh
= 2 × (22/7) × 0.25 × 3.5
= 5.5 m²

Cost of painting 1 m² = ₹ 12.50
∴ Total cost of painting the curved surface of the pillar
= ₹ 12.50 × 5.5 = ₹ 68.75 (Answer)



Q6: Curved surface area of a right circular cylinder is 4.4 m². If the radius of the base of the cylinder is 0.7 m, find its height.

Answer: Given,
CSA of the cylinder = 4.4 m²,
r = 0.7 m,
h = ?
CSA = 2πrh
⇒ 4.4 = 2 × (22/7) × 0.7 × h
⇒ h = 4.4/4.4 = 1m (Answer)



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 Rs 40 per m².

Answer: Given, r = 3.5/2 = 1.75 m,

h = 10m


(i) Inner curved surface area of the well

= 2πrh = 2 × (22/7) × 1.75 × 10

= 22 × 5

= 110 m² (Answer)


(ii) Cost of plastering 1 m² = ₹ 40

∴ Cost of plastering the curved surface area of the well

= ₹ 110 × 40 = ₹ 4400 (Answer)

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.

Answer: Given, r = 5/2 = 2.5 cm = 0.025m, h = 28m

Total radiating surface in the system = total surface area of the cylinder

= 2π r(h + r)

= 2 ×  (22/7) ×  (0.025) × (0.025 + 28)

= (44 × 0.025 ×  28.025)/7 = 4.403 m² (Answer)

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 1/12th of the steel actually used was wasted in making the tank.


Answer: Here, r = 4.2/2 = 2.1 m, h = 4.5 m
(i) CSA of the storage tank = 2πrh
= 2 × (22/7) × 2.1 × 4.5 m²
= 59.4 m² (Answer)


(ii) Total surface area of the tank = 2πr (h + r)
= 2 × (22/7) × 2.1 (4.5 + 2.1)
= 44 × 0.3 × 6.6 m2 = 87.12 m²
Let the actual area of steel used be x m².
Area of steel wasted = (1/12) of x = (x/12) m².
∴ area of the steel used in the tank = x - (x/12) = (11x/12)
⇒ 87.12 = (11x/12)
⇒ x = 87.12 × 12/11 =  95.04 m²

Thus, 95.04 m² of steel was actually used.

Q10: In the figure, 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.

Answer: Given, r =20/2 = 10cm, h = 30 cm
Circumference of the frame = 2πr
= 2π × 10 cm = 20π cm
Height of the frame = 30 cm
Height of the cloth needed to cover the frame (including margin) = 30 + 2.5 + 2.5 = 35cm
Also, breadth of the cloth = circumference of the base of the frame.
∴ Area of the cloth required for covering the lampshade = length × breadth
= 35 × 20π
= 35 × 20 × (22/7)
= 2200 cm²


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?

Answer:
Given,  r = 3 cm, h = 10.5 cm
The penholders have only one base i.e., these are open at one end.
Total surface area (TSA) of 1 penholder = 2πrh + πr²
= πr (2h + + r)
= (22/7) × 3 (2 × 10.5 + 3)
= (22/7) × 3 × 24
TSA for 35 pen holders = (22/7) × 3 × 24 × 35

= 7920 cm² (Answer)