GSEB Class 9 Maths Solutions Chapter 13 Surface Areas and Volumes 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: Let r cm be the radius of the cylinder's base.
The height \( (h) \) is 14 cm.
The curved surface area \( (\text{CSA}) \) measures 88 cm².
The formula for curved surface area is \( 2 \pi r h \).
So, \( 2 \pi r h = 88 \)
\( 2 \times \frac{22}{7} \times r \times 14 = 88 \)
\( 2 \times 22 \times r \times 2 = 88 \)
\( 88r = 88 \)
\( r = \frac{88}{88} \)
\( r = 1 \text{ cm} \)
The diameter of the base is \( 2r \).
\( \text{Diameter} = 2 \times 1 = 2 \text{ cm} \)
Therefore, the base diameter of the cylinder is 2 cm.
In simple words: The cylinder's side area is 88 cm² and it's 14 cm tall. We used the formula for curved area to find its radius, which came out to be 1 cm. Then, we doubled the radius to get the diameter, which is 2 cm.

Exam Tip: Always remember the formula for the curved surface area of a cylinder and ensure consistent units throughout your calculations. Clearly state your final answer with the correct unit.

Question 2. It is required to make a closed cylindrical tank of height 1 m and a base diameter 140 cm sheet. How many square meters of the sheet is required for the same?
Answer: The height of the tank \( (h) = 1 \text{ m} = 100 \text{ cm} \).
The base diameter \( (2r) = 140 \text{ cm} \).
So, the radius \( (r) = \frac{140}{2} \text{ cm} = 70 \text{ cm} \).
The total surface area of a closed cylindrical tank is given by \( 2 \pi r (h + r) \).
\( = 2 \times \frac{22}{7} \times 70 \times (100 + 70) \)
\( = 2 \times 22 \times 10 \times 170 \)
\( = 440 \times 170 \)
\( = 74800 \text{ cm}^2 \)
To convert cm² to m², divide by \( 100 \times 100 \).
\( = \frac{74800}{100 \times 100} \text{ m}^2 = \frac{74800}{10000} \text{ m}^2 = 7.48 \text{ m}^2 \)
Therefore, 7.48 square meters of the sheet are necessary.
In simple words: We need to build a closed cylinder with a height of 1 m and a base diameter of 140 cm. First, we found the total surface area in cm² using the right formula. Then, we changed that area to square meters.

Exam Tip: When dealing with cylindrical tanks, pay close attention to whether the tank is open or closed, as this affects the surface area formula. Always convert all units to be consistent (e.g., all meters or all centimeters) before beginning calculations, and convert back to the required unit for the final answer.

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. Find its
1. inner curved surface area,
2. outer curved surface area,
3. the total surface area.
Answer: The length (height) of the metal pipe \( (h) = 77 \text{ cm} \).
The inner diameter \( (2r) = 4 \text{ cm} \), so the inner radius \( (r) = 2 \text{ cm} \).
The outer diameter \( (2R) = 4.4 \text{ cm} \), so the outer radius \( (R) = 2.2 \text{ cm} \).
1. Inner curved surface area:
\( = 2 \pi r h \)
\( = 2 \times \frac{22}{7} \times 2 \times 77 \)
\( = 2 \times 22 \times 2 \times 11 \)
\( = 968 \text{ cm}^2 \)
2. Outer curved surface area:
\( = 2 \pi R h \)
\( = 2 \times \frac{22}{7} \times 2.2 \times 77 \)
\( = 2 \times 22 \times 2.2 \times 11 \)
\( = 1064.8 \text{ cm}^2 \)
3. Total surface area:
The total surface area of a hollow cylinder includes inner curved surface area, outer curved surface area, and the areas of the two circular rings (top and bottom).
Total surface area \( = 2 \pi R h + 2 \pi r h + 2 \pi (R^2 - r^2) \)
\( = 1064.8 + 968 + 2 \times \frac{22}{7} [(2.2)^2 - (2)^2] \)
\( = 1064.8 + 968 + 2 \times \frac{22}{7} [4.84 - 4] \)
\( = 1064.8 + 968 + 2 \times \frac{22}{7} \times 0.84 \)
\( = 1064.8 + 968 + 2 \times 22 \times 0.12 \)
\( = 1064.8 + 968 + 5.28 \)
\( = 2038.08 \text{ cm}^2 \)
In simple words: First, we calculated the inner curved surface area of the pipe and then the outer curved surface area. After that, we found the area of the two ring-shaped ends (top and bottom). Finally, we added all these areas together to get the pipe's total surface area.

Exam Tip: For hollow cylinders or pipes, remember that the total surface area involves three parts: the inner curved surface, the outer curved surface, and the area of the two annular (ring-shaped) ends. Make sure to use the correct radii for each calculation.

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: The diameter of the roller \( (2r) = 84 \text{ cm} \).
So, the radius \( (r) = \frac{84}{2} = 42 \text{ cm} \).
The length (height) of the roller \( (h) = 120 \text{ cm} \).
The area covered by the roller in 1 complete revolution is its curved surface area.
Area in 1 revolution \( = 2 \pi r h \)
\( = 2 \times \frac{22}{7} \times 42 \times 120 \)
\( = 2 \times 22 \times 6 \times 120 \)
\( = 31680 \text{ cm}^2 \)
The roller makes 500 complete revolutions to level the playground.
Total area of the playground \( = 31680 \times 500 \)
\( = 15840000 \text{ cm}^2 \)
To convert cm² to m², we divide by \( 100 \times 100 \).
\( = \frac{15840000}{100 \times 100} \text{ m}^2 = \frac{15840000}{10000} \text{ m}^2 = 1584 \text{ m}^2 \)
Therefore, the playground's area is 1584 square meters.
In simple words: The roller's curved surface area tells us how much ground it covers in one spin. We multiplied this by 500 (the number of spins) to get the total playground area in square centimeters. Finally, we changed the area from square centimeters to square meters.

Exam Tip: Remember that the area covered by a roller in one revolution is equal to its curved surface area. Be careful with unit conversions, especially from cm² to m², as it requires dividing by \( 100^2 \).

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 ₹ 12.50 per m².
Answer: The diameter of the cylindrical pillar \( (2r) = 50 \text{ cm} \).
So, the radius \( (r) = \frac{50}{2} = 25 \text{ cm} \).
Convert the radius to meters: \( r = 0.25 \text{ m} \).
The height of the pillar \( (h) = 3.5 \text{ m} \).
The curved surface area of the pillar is \( 2 \pi r h \).
\( = 2 \times \frac{22}{7} \times 0.25 \times 3.5 \)
\( = 2 \times \frac{22}{7} \times \frac{1}{4} \times \frac{7}{2} \)
\( = \frac{22}{4} \)
\( = 5.5 \text{ m}^2 \)
The rate of painting is Rs 12.50 per m².
The expense for painting the curved surface area of the pillar \( = 5.5 \times 12.50 \)
\( = \text{Rs } 68.75 \)
In simple words: We first converted the pillar's diameter to meters to match the height. Then, we calculated the curved surface area. Finally, we multiplied this area by the given cost per square meter to find the total painting expense.

Exam Tip: Always make sure all dimensions are in the same units before starting calculations. The curved surface area is the part to be painted, not the total surface area unless specified otherwise.

Question 6. The 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: Let h be the height of the right circular cylinder, measured in meters.
The radius of the base \( (r) = 0.7 \text{ m} \).
The curved surface area \( (\text{CSA}) = 4.4 \text{ m}^2 \).
The formula for the curved surface area of a cylinder is \( 2 \pi r h \).
So, \( 2 \pi r h = 4.4 \)
\( 2 \times \frac{22}{7} \times 0.7 \times h = 4.4 \)
\( 2 \times 22 \times 0.1 \times h = 4.4 \)
\( 4.4 h = 4.4 \)
\( h = \frac{4.4}{4.4} \)
\( h = 1 \text{ m} \)
Therefore, the right circular cylinder's height is 1 meter.
In simple words: We used the given curved surface area and radius to work backwards and find the cylinder's height. By plugging the known values into the curved surface area formula, we solved for the unknown height.

Exam Tip: When an unknown dimension (like height or radius) needs to be found, set up the appropriate surface area formula with the given values and solve the equation. Double-check your algebraic manipulation to avoid errors.

Question 7. The inner diameter of a circular wall is 3.5 m. It is 10 m deep. Find
1 its inner curved surface area,
2. the cost of plastering this curved surface at the rate of ₹ 40 per m².
Answer: The inner diameter of the circular wall \( (2r) = 3.5 \text{ m} \).
So, the inner radius \( (r) = \frac{3.5}{2} \text{ m} = 1.75 \text{ m} \).
The depth of the wall (which acts as height) \( (h) = 10 \text{ m} \).
1. Inner curved surface area:
\( = 2 \pi r h \)
\( = 2 \times \frac{22}{7} \times 1.75 \times 10 \)
\( = 2 \times \frac{22}{7} \times \frac{7}{4} \times 10 \)
\( = 22 \times 5 \)
\( = 110 \text{ m}^2 \)
2. The expense for plastering the curved surface, given the rate of Rs 40 per square meter:
\( = 110 \times 40 \)
\( = \text{Rs } 4400 \)
In simple words: We calculated the inner curved surface area of the circular wall first. Then, we multiplied this area by the cost per square meter to find out the total amount of money needed for plastering.

Exam Tip: For problems involving wells or circular walls, the 'depth' typically refers to the height of the cylindrical part. Remember to calculate the curved surface area before determining the total cost of plastering or painting.

Question 8. In a hot water heating system, there is a cylindrical pipe of length 28 m and a diameter 5 cm. Find the total radiating surface in the system.
Answer: The length (height) of the cylindrical pipe \( (h) = 28 \text{ m} \).
The diameter \( (2r) = 5 \text{ cm} \).
So, the radius \( (r) = \frac{5}{2} \text{ cm} \).
Convert radius to meters: \( r = \frac{5}{2 \times 100} \text{ m} = \frac{5}{200} \text{ m} = \frac{1}{40} \text{ m} \).
The total radiating surface in the system is the curved surface area of the pipe.
Total radiating surface \( = 2 \pi r h \)
\( = 2 \times \frac{22}{7} \times \frac{1}{40} \times 28 \)
\( = 2 \times 22 \times \frac{1}{40} \times 4 \)
\( = \frac{2 \times 22 \times 4}{40} \)
\( = \frac{176}{40} \)
\( = 4.4 \text{ m}^2 \)
In simple words: The pipe's length is 28 m and its diameter is 5 cm. We converted the diameter to meters and then found the curved surface area of the pipe, as this is the radiating surface. The final area is 4.4 square meters.

Exam Tip: When different units are provided (e.g., length in meters, diameter in centimeters), always convert them to a single consistent unit before performing calculations. The "radiating surface" of a pipe typically refers to its curved surface area.

Question 9. Find:
1. 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.
2. how much steel was actually used, if \( \frac{1}{12} \) of the steel actually used was wasted in making the tank.
Answer: 1. For the cylindrical petrol storage tank:
Diameter \( (2r) = 4.2 \text{ m} \).
So, radius \( (r) = \frac{4.2}{2} \text{ m} = 2.1 \text{ m} \).
Height \( (h) = 4.5 \text{ m} \).
The lateral or curved surface area \( = 2 \pi r h \)
\( = 2 \times \frac{22}{7} \times 2.1 \times 4.5 \)
\( = 2 \times 22 \times 0.3 \times 4.5 \)
\( = 44 \times 1.35 \)
\( = 59.4 \text{ m}^2 \)
2. To find the total surface area of the closed cylindrical tank:
Total surface area \( = 2 \pi r (h + r) \)
\( = 2 \times \frac{22}{7} \times 2.1 \times (4.5 + 2.1) \)
\( = 2 \times \frac{22}{7} \times 2.1 \times 6.6 \)
\( = 2 \times 22 \times 0.3 \times 6.6 \)
\( = 44 \times 1.98 \)
\( = 87.12 \text{ m}^2 \)
Let x m² represent the actual steel area used.
Since \( \frac{1}{12} \) of the actual steel was wasted, the portion of steel that formed the tank is \( (1 - \frac{1}{12}) \text{ of } x = \frac{11}{12} \text{ of } x \).
This area must equal the total surface area of the tank.
So, \( \frac{11}{12} x = 87.12 \)
\( x = \frac{87.12 \times 12}{11} \)
\( x = 7.92 \times 12 \)
\( x = 95.04 \text{ m}^2 \)
The total steel actually used amounts to 95.04 square meters.
In simple words: First, we calculated the curved surface area of the petrol tank. Then, we found its total surface area. Since some steel was wasted, we determined that the tank's surface area represents only 11/12 of the steel actually bought. From this, we worked out the total amount of steel that was purchased.

Exam Tip: Carefully distinguish between lateral/curved surface area and total surface area based on the question. For problems involving wasted material, remember that the calculated surface area of the final object is the usable portion of the original material. Set up an equation to find the total original material needed.

Question 10. In the figure given below, 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: The base diameter of the lampshade frame \( (2r) = 20 \text{ cm} \).
So, the radius \( (r) = \frac{20}{2} = 10 \text{ cm} \).
The original height of the frame \( (h) = 30 \text{ cm} \).
A margin of 2.5 cm is needed for folding at the top and another 2.5 cm for the bottom.
Total margin \( = 2.5 \text{ cm} + 2.5 \text{ cm} = 5 \text{ cm} \).
The total height of the cloth required \( (h') = h + \text{total margin} \)
\( h' = 30 \text{ cm} + 5 \text{ cm} = 35 \text{ cm} \).
The cloth needed to cover the lampshade (which is the curved surface area) \( = 2 \pi r h' \)
\( = 2 \times \frac{22}{7} \times 10 \times 35 \)
\( = 2 \times 22 \times 10 \times 5 \)
\( = 2200 \text{ cm}^2 \)
In simple words: We added the extra margin needed for folding to the lampshade's height. Then, we calculated the curved surface area using this new, taller height and the given radius to find the total amount of cloth required.

Exam Tip: When a margin is added for folding, it increases the effective height of the material needed, usually at both the top and bottom. Remember to add the margin twice to the original height (once for each end) before calculating the surface area.

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: For each penholder (a cylinder with a base and an open top):
Radius \( (r) = 3 \text{ cm} \).
Height \( (h) = 10.5 \text{ cm} \).
The cardboard required for one penholder (curved surface area + base area) \( = 2 \pi r h + \pi r^2 \)
\( = (2 \times \frac{22}{7} \times 3 \times 10.5) + (\frac{22}{7} \times 3^2) \)
\( = (2 \times \frac{22}{7} \times 3 \times \frac{21}{2}) + (\frac{22}{7} \times 9) \)
\( = (22 \times 3 \times 3) + (\frac{198}{7}) \)
\( = 198 + \frac{198}{7} \)
\( = 198 (1 + \frac{1}{7}) \)
\( = 198 \times \frac{8}{7} \text{ cm}^2 \)
There are 35 competitors.
Total cardboard required for 35 competitors \( = (198 \times \frac{8}{7}) \times 35 \)
\( = 198 \times 8 \times 5 \)
\( = 198 \times 40 \)
\( = 7920 \text{ cm}^2 \)
Therefore, 7920 square centimeters of cardboard needed to be bought for the competition.
In simple words: Each penholder needed cardboard for its curved side and its base. We calculated this area for one penholder. Since 35 students were making them, we multiplied that single penholder's area by 35 to find the total cardboard required.

Exam Tip: When an object has an open top (like a penholder), its total surface area does not include the top circular area. The formula should be \( 2 \pi r h + \pi r^2 \). Multiply the area for one item by the number of items to get the total material required.