Maharashtra Board Class 9 Maths Part 2 Geometry Chapter 9 Set 9.3 Surface Area and Volume Solutions

Question 1. Find the surface areas and volumes of spheres of the following radii:
i. 4 cm
ii. 9 cm
iii. 3.5 cm (\( \pi = 3.14 \))
Answer:
i. Radius (\( r \)) = 4 cm
Surface Area of sphere = \( 4\pi r^2 \)
\( \implies \) Surface Area = \( 4 \times 3.14 \times 4^2 \)
\( \implies \) Surface Area = \( 4 \times 3.14 \times 16 \)
\( \implies \) Surface Area = \( 200.96\text{ cm}^2 \)

Volume of sphere = \( \frac{4}{3}\pi r^3 \)
\( \implies \) Volume = \( \frac{4}{3} \times 3.14 \times 4^3 \)
\( \implies \) Volume = \( \frac{4}{3} \times 3.14 \times 64 \)
\( \implies \) Volume = \( \frac{803.84}{3} \approx 267.95\text{ cm}^3 \)

ii. Radius (\( r \)) = 9 cm
Surface Area of sphere = \( 4\pi r^2 \)
\( \implies \) Surface Area = \( 4 \times 3.14 \times 9^2 \)
\( \implies \) Surface Area = \( 4 \times 3.14 \times 81 \)
\( \implies \) Surface Area = \( 1017.36\text{ cm}^2 \)

Volume of sphere = \( \frac{4}{3}\pi r^3 \)
\( \implies \) Volume = \( \frac{4}{3} \times 3.14 \times 9^3 \)
\( \implies \) Volume = \( \frac{4}{3} \times 3.14 \times 729 \)
\( \implies \) Volume = \( 4 \times 3.14 \times 243 \)
\( \implies \) Volume = \( 3052.08\text{ cm}^3 \)

iii. Radius (\( r \)) = 3.5 cm (\( \pi = 3.14 \))
Surface Area of sphere = \( 4\pi r^2 \)
\( \implies \) Surface Area = \( 4 \times 3.14 \times (3.5)^2 \)
\( \implies \) Surface Area = \( 4 \times 3.14 \times 12.25 \)
\( \implies \) Surface Area = \( 153.86\text{ cm}^2 \)

Volume of sphere = \( \frac{4}{3}\pi r^3 \)
\( \implies \) Volume = \( \frac{4}{3} \times 3.14 \times (3.5)^3 \)
\( \implies \) Volume = \( \frac{4}{3} \times 3.14 \times 42.875 \)
\( \implies \) Volume = \( \frac{538.51}{3} \approx 179.50\text{ cm}^3 \). Calculating these values carefully helps us understand how the size of a sphere affects its overall space and capacity.
In simple words: To find the surface area and volume of a sphere, we use standard formulas with the given radius. Surface area measures the outer boundary, while volume measures the total space inside the sphere.

🎯 Exam Tip: Always write down the general formulas for surface area and volume before substituting the values, and remember to include the correct units like \(\text{cm}^2\) for area and \(\text{cm}^3\) for volume to secure full marks.

Question 1. Find the surface area and volume of a sphere with the following radii:
(i) 4 cm
(ii) 9 cm
(iii) 3.5 cm
Answer:
(i) Given: Radius \( (r) = 4\text{ cm} \)
To find: Surface area and volume of sphere
Solution:
Surface area of sphere = \( 4\pi r^2 \)
\( = 4 \times 3.14 \times 4^2 \)
\( \therefore \) Surface area of sphere = \( 200.96\text{ sq.cm} \)
Volume of sphere = \( \frac{4}{3}\pi r^3 \)
\( = \frac{4}{3} \times 3.14 \times 4^3 \)
\( \therefore \) Volume of sphere = \( 267.95\text{ cubic cm} \)

(ii) Given: Radius \( (r) = 9\text{ cm} \)
To find: Surface area and volume of sphere
Solution:
Surface area of sphere = \( 4\pi r^2 \)
\( = 4 \times 3.14 \times 9^2 \)
\( \therefore \) Surface area of sphere = \( 1017.36\text{ sq.cm} \)
Volume of sphere = \( \frac{4}{3}\pi r^3 \)
\( = \frac{4}{3} \times 3.14 \times 9^3 \)
\( = \frac{4}{3} \times 3.14 \times 9 \times 9 \times 9 \)
\( = 4 \times 3.14 \times 3 \times 9 \times 9 \)
\( \therefore \) Volume of sphere = \( 3052.08\text{ cubic cm} \)

(iii) Given: Radius \( (r) = 3.5\text{ cm} \)
To find: Surface area and volume of sphere
Solution:
Surface area of sphere = \( 4\pi r^2 \)
\( = 4 \times 3.14 \times (3.5)^2 \)
\( \therefore \) Surface area of sphere = \( 153.86\text{ sq.cm} \)
Volume of sphere = \( \frac{4}{3}\pi r^3 \)
\( = \frac{4}{3} \times 3.14 \times (3.5)^3 \)
\( \therefore \) Volume of sphere = \( 179.50\text{ cubic cm} \)
In simple words: To find the surface area and volume of a sphere, we use standard formulas that depend only on the radius. By plugging in the given radius and the value of pi, we can easily calculate both values.

🎯 Exam Tip: Always write down the formula clearly before substituting the values to secure step-marks. Double-check your calculations, especially when dealing with decimals and exponents.

Question 2. If the radius of a solid hemisphere is 5 cm, then find its curved surface area and total surface area. (\( \pi = 3.14 \))
Answer:
Given: Radius \( (r) = 5\text{ cm} \)
To find: Curved surface area and total surface area of the solid hemisphere
Curved surface area of hemisphere = \( 2\pi r^2 \)
\( = 2 \times 3.14 \times 5^2 \)
\( = 2 \times 3.14 \times 25 \)
\( = 50 \times 3.14 \)
\( \therefore \) Curved surface area of hemisphere = \( 157\text{ sq.cm} \)

Total surface area of solid hemisphere = \( 3\pi r^2 \)
\( = 3 \times 3.14 \times 5^2 \)
\( = 3 \times 3.14 \times 25 \)
\( = 75 \times 3.14 \)
\( \therefore \) Total surface area of solid hemisphere = \( 235.5\text{ sq.cm} \)
In simple words: A hemisphere is half of a sphere, so its curved surface area is half of a sphere's surface area. Its total surface area also includes the flat circular base on the bottom, which adds another \( \pi r^2 \) to the total.

🎯 Exam Tip: Remember that the total surface area of a solid hemisphere includes the base area (\( \pi r^2 \)), making it \( 3\pi r^2 \) instead of just \( 2\pi r^2 \).

Question 3. If the surface area of a sphere is \( 2826 \text{ cm}^2 \) then find its volume. (\( \pi = 3.14 \))
Answer:
Given: Surface area of sphere = \( 2826 \text{ sq. cm.} \)
To find: Volume of sphere
i. Surface area of sphere = \( 4\pi r^2 \)
\( \therefore 2826 = 4 \times 3.14 \times r^2 \)
\( \therefore r^2 = \frac{2826}{4 \times 3.14} = \frac{282600}{4 \times 314} = \frac{900}{4} \)
\( \dots r^2 = 225 \)
\( \therefore r = \sqrt{225} \) ... [Taking square root on both sides]
\( = 15 \text{ cm} \)

ii. Volume of sphere = \( \frac{4}{3}\pi r^3 \)
\( = \frac{4}{3} \times 3.14 \times 15^3 \)
\( = \frac{4}{3} \times 3.14 \times 15 \times 15 \times 15 \)
\( = 4 \times 3.14 \times 5 \times 15 \times 15 \)
\( = 14130 \text{ cubic cm.} \)
\( \dots \) The volume of the sphere is \( 14130 \text{ cubic cm.} \) This calculation helps us understand the total space enclosed within the sphere.
In simple words: First, we use the given surface area to find the radius of the sphere, which is 15 cm. Then, we use this radius in the volume formula to find that the sphere can hold 14130 cubic cm of space.

🎯 Exam Tip: When solving sphere problems, always find the radius \( r \) first, as it is the key variable needed for both surface area and volume formulas.

Question 4. Find the surface area of a sphere, if its volume is \( 38808 \text{ cubic cm.} \) (\( \pi = \frac{22}{7} \))
Answer:
Given: Volume of sphere = \( 38808 \text{ cubic cm.} \)
To find: Surface area of sphere
Volume of sphere = \( \frac{4}{3} \pi r^3 \)
\( \therefore 38808 = \frac{4}{3} \times \frac{22}{7} \times r^3 \)
\( \therefore r^3 = \frac{38808 \times 3 \times 7}{4 \times 22} \)
\( \dots r^3 = 441 \times 21 \)
\( \therefore r^3 = 9261 \)
\( \therefore r = 21 \text{ cm} \) ... [Taking cube root on both sides]

Surface area of sphere = \( 4 \pi r^2 \)
\( = 4 \times \frac{22}{7} \times 21 \times 21 \)
\( = 4 \times 22 \times 3 \times 21 \)
\( = 5544 \text{ sq. cm.} \)
\( \therefore \) The surface area of the sphere is \( 5544 \text{ sq. cm.} \) This represents the total outer boundary area of the given sphere.
In simple words: We first use the volume formula to find that the radius of the sphere is 21 cm. Then, we plug this radius into the surface area formula to find that the outer surface area is 5544 sq. cm.

🎯 Exam Tip: Be careful with calculations involving large numbers; simplify by dividing step-by-step before multiplying to avoid arithmetic errors.

Question 4. Find the surface area of a sphere if its volume is 38808 cubic cm.
Answer:
i. Volume of sphere = \( \frac{4}{3}\pi r^3 \)
\( \therefore 38808 = \frac{4}{3} \times \frac{22}{7} \times r^3 \)
\( \therefore r^3 = \frac{38808 \times 3 \times 7}{4 \times 22} \)
\( = \frac{9702 \times 3 \times 7}{22} \)
\( \therefore r^3 = 441 \times 21 = 21 \times 21 \times 21 \)
\( \therefore r = 21 \text{ cm} \) ... [Taking cube root on both sides]
ii. Surface area of sphere = \( 4\pi r^2 \)
\( = 4 \times \frac{22}{7} \times 21 \)
\( = 4 \times \frac{22}{7} \times 21 \times 21 \)
\( = 4 \times 22 \times 3 \times 21 \)
\( = 5544 \text{ sq.cm.} \)
\( \therefore \) The surface area of sphere is \( 5544 \text{ sq.cm.} \) This calculation shows how the volume of a three-dimensional curved shape directly relates to its outer boundary area.
In simple words: To find the surface area, we first used the given volume to calculate the radius of the sphere. Once we found the radius to be 21 cm, we plugged it into the surface area formula to get the final answer.

🎯 Exam Tip: Remember to write the correct units, such as cm for radius and sq.cm. for surface area, to avoid losing marks.

Question 5. Volume of a hemisphere is 18000\(\pi\) cubic cm. Find its diameter.
Answer:
Given: Volume of hemisphere = \( 18000\pi \text{ cubic cm.} \)
To find: Diameter of the hemisphere
i. Volume of hemisphere = \( \frac{2}{3}\pi r^3 \)
\( \therefore 18000\pi = \frac{2}{3}\pi r^3 \)
\( \dots 18000 = \frac{2}{3}r^3 \)
\( \therefore r^3 = \frac{18000 \times 3}{2} \)
\( = 9000 \times 3 \)
\( \therefore r^3 = 27000 \)
\( \dots r = \sqrt[3]{27000} \) ... [Taking cube root on both sides]
\( = 30 \text{ cm} \)
ii. Diameter = \( 2r \)
\( = 2 \times 30 = 60 \text{ cm} \)
\( \therefore \) The diameter of the hemisphere is \( 60 \text{ cm.} \) This relationship helps us easily find the width of a curved bowl-like shape when we know how much space it encloses.
In simple words: We used the volume of the hemisphere to find its radius, which came out to be 30 cm. Since the diameter is always double the radius, we multiplied 30 by 2 to get 60 cm.

🎯 Exam Tip: When dealing with terms containing \( \pi \) on both sides, cancel them out early to simplify your calculations quickly.