13. Surface Areas and Volumes Mathematics Exercise - 13.2 class 9 Maths in English - CBSE Study
NCERT Solutions for Class 9 Mathematics are carefully prepared according to the latest CBSE syllabus and NCERT textbooks to help students understand every concept clearly. These solutions cover all important 13. Surface Areas and Volumes with detailed explanations and step-by-step answers for better exam preparation. Each Exercise 13.2 is explained in simple language so that students can easily grasp the fundamentals and improve their academic performance. The study material is designed to support daily homework, revision practice, and final exam preparation for Class 9 students. With accurate answers, concept clarity, and structured content, these NCERT solutions help learners build confidence and score higher marks in their examinations. Whether you are revising a specific topic or preparing an entire chapter, this resource provides reliable and syllabus-based guidance for complete success in Mathematics.
Class 9 English Medium Mathematics All Chapters:
13. Surface Areas and Volumes
2. 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 cm2. Find the diameter of the base of the cylinder.
Solution:
H = 14cm, curved surface area of cylinder = 88cm2
Curved surface area of cylinder = 88cm2
⇒ 2 × × r × h = 88cm2
⇒ 2 × × r × 14 = 88cm2
⇒ 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
⇒ 7.48m2
Hence, 7.48m2 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 cm2
(ii) exterior curved surface area = 2πrh
⇒ 5.28 cm2
Total surface area = interior surface area + exterior surface area + ring surface area
⇒ 1064.8 + 9.68 + 5.28 cm2
⇒ 2038.08 cm2
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.
Solution:
D = 84cm ⇒ r = 42cm, H = 120cm
Curved surface area of cylinder = 2πrh
⇒ 2 × × 42 × 120
⇒ 44 × 6 × 120
⇒ 31680 cm2
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 m2.
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 m2. If the radius of the base of the cylinder is 0.7 m, find its height.
Solution:
Curved surface area of cylinder = 4.4 m2 , 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 m2.
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 + πr2
⇒ πr(2h + r)
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