Class 9 - Mathematics
Surface Areas and Volumes - Exercise 13.2
Top Block 1
Exercise 13.2
Question :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.
Answer :
Here, h = 14 cm, curved, surface area = 88 cm2, r = ?
Curved surface area of the cylinder = 2πrh
⇒ 88 = 2 * (22/7) * r * 14
⇒ 88 = 44 × 2 × r
⇒ 88 = 88 * r
⇒ r = 88/88
⇒ r = 1 cm
Hence, the base diameter of cylinder = 1 * 2 = 2 cm
Question : 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 are required for the same?
Answer :
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)
= 44 * 0.1 * 1.7
= 7.48 m2
Hence, 7.48 m2 of sheet is required.
Question : 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.
Answer :
Here, h = 77 cm,
Outer radius (R) = 4.4/2 cm = 2.2 cm
Inner radius (r) = 4/2 cm = 2 cm
(i) Inner curved surface area of the pipe = 2πrh
= 2 * (22/7) * 2 * 77
= 2 * 22 * 22
= 968 cm2
(ii) Outer curved surface area of the pipe = 2πRh
= 2 * (22/7) * 2.2 * 77
= 44 * 24.2
= 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 + 1064.80 + 2 * (22/7)[(2.2)2 – 22]
= 2032.80 + 5.28
= 2038.08 cm2
Question : 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.
Answer :
Radius of the roller (r) = 84/2 = 42 cm
Length of the roller (h) = 120 cm
Curved surface area of the roller = 2πrh
= 2 * (22/7) * 42 * 120
= 44 * 6 * 120
= 44 * 720
= 31680 cm2
Since area covered by the roller in 1 revolution = 31680 cm2
So, area covered by the roller in 500 revolutions = 31680 × 500 cm2
= 15840000 cm2
Hence, area of the playground = 15840000/(100 * 100)
= 15840000/10000
= 1584 cm2
Question : 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.
Answer :
Here, r = 50/2 = 25 cm = 0.25 m, h = 3.5 m
Curved surface area of the pillar = 2πrh
= 2 * (22/7) * 0.25 * 3.5
= 44 * 0.25 * 0.5
= 5.5 m2
Cost of painting 1 m2 = Rs 12.50
So, total cost of painting the curved surface of the pillar = Rs 12.50 * 5.5 = Rs 68.75
Question : 6: 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 :
Curved surface area of the cylinder = 4.4 m2, r = 0.7 m, h = ?
Curved surface area of the cylinder = 2πrh
⇒ 4.4 = 2 * (22/7) * 0.7 * h
⇒ 4.4 = 44 * 0.1 * h
⇒ 4.4 = 4.4 * h
⇒ h = 4.4/4.4
⇒ h = 1
Hence, height of the cylinder is 1 m.
Question : 7: 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 :
Here, r = 3.5/2 m , h = 10 m
(i) Inner curved surface area of the well = 2πrh
= 2 * (22/7) * (3.5/2) * 10
= 22 * 5 = 110 m2
(ii) Cost of plastering 1 m2 = Rs 40
So, Cost of plastering the curved surface area of the well = Rs 110 * 40 = Rs 4400
Question : 8 :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 :
Here, r = 5/2 cm = 2.5 cm = 0.025 m, h = 28 m.
Total radiating surface in the system = total surface area of the cylinder
= 2πr(h + r)
= 2 * (22/7) * 0.025 (28 + 0.025)
= (44 * 0.025 * 28.025)/7
= 4.4 m2 (approx)
Question : 9: 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 1/12 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) Curved surface area of the storage tank = 2πrh
= 2 * (22/7) * 2.1 * 4.5
= 44 * 0.3 * 4.5
= 59.4 m2
(ii) Total surface area of the tank = 2πr(h + r)
= 2 * (22/7) * 2.1 4.5 + 2.1)
= 44 * 0.3 * 6.6
= 87.12 m2
Let the actual area of steel used be x m2.
Area of steel wasted = 1/12 of x
= x/12 m2
So, area of the steel used in the tank = (x – x/12) = 11x/12 m2
⇒ 87.12 = 11x/12
⇒ x = (87.12 * 12)/11
⇒ x = 7.92 * 12
⇒ x = 95.04 m2
Hence, 95.04 m2 of steel was actually used.
Question : 10: 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.
Mddle block 1
Answer :
Here, r = 20/2 cm = 10 cm, Height = 30 cm
Circumference of the base of the frame = 2πr
= 2π * 10 = 20π cm
Height of the frame = 30 cm
Height of the cloth needed for covering the frame (including the margin)
= (30 + 2.5 + 2.5) cm = 35 cm
Also, breadth of the cloth = circumference of the base of the frame.
So, area of the cloth required for covering the lampshade = length * breadth
= 35 * 20π
= 35 * 20 * (22/7)
= 5 * 20 * 22
= 2200 cm2
Question : 11: 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 :
Here, r = 3 cm, h = 10.5 cm
The penholders have only one base i.e., these are open at one end.
Total surface area of 1 penholder = 2πrh + πr2
= πr(2h + r)
= (22/7) * 3 (2 * 10.5 + 3)
= (22/7) * 3 * 24
Total surface area of 35 penholders = (22/7) * 3 * 24 * 35
= 22 * 3 * 24 * 5
= 7920 cm2
Hence, 7920 cm2 of cardboard is needed.
