Important Questions for Class 9 Maths (Ganita Manjari II) Chapter 14 Math of Space: Surface Area and Volume 2026-27

Understand surface area and volume through solved questions with clear steps. Download the PDF for offline practice and use the answers to check your calculations.

Class 9 Maths Chapter 14 MCQs with Answers

1. The volume of a cube is 64 cm³. What is its total surface area?

A. 16 cm²

B. 64 cm²

C. 96 cm²

D. 128 cm²

Answer: C. 96 cm².

Solution:

Let the side of the cube be a.

$$a^3=64$$

Therefore, a = 4 cm.

$$\text{TSA}=6a^2=6\times4^2=96\text{ cm}^2$$

2. What shape is obtained when the curved surface of a right circular cylinder is cut along its height and unrolled?

A. Triangle

B. Rectangle

C. Circle

D. Sector of a circle

Answer: B. Rectangle.

3. A cone and a cylinder have the same base radius and height. What is the ratio of their volumes?

A. 1 : 2

B. 1 : 3

C. 2 : 3

D. 3 : 1

Answer: B. 1 : 3.

Solution:

$$V_{\text{cone}}=\dfrac{1}{3}\pi r^2h$$

$$V_{\text{cylinder}}=\pi r^2h$$

Therefore, their volumes are in the ratio 1 : 3.

4. Which expression gives the total surface area of a solid hemisphere of radius r?

A. $$2\pi r^2$$

B. $$3\pi r^2$$

C. $$4\pi r^2$$

D. $$\dfrac{2}{3}\pi r^3$$

Answer: B. $$3\pi r^2$$.

5. The length of a cuboid is increased by 1 unit while its width w and height h remain unchanged. What is the increase in its volume?

A. 1 cubic unit

B. lw cubic units

C. wh cubic units

D. lh cubic units

Answer: C. wh cubic units.

Solution:

$$\text{Increase}=(l+1)wh-lwh=wh$$

6. Why does dividing a room’s volume by the volume of one tennis ball not give the exact number of balls that fit inside it?

A. Tennis balls have no fixed volume.

B. Gaps remain between the balls.

C. The room must be cylindrical.

D. Surface area determines the number of balls.

Answer: B. Gaps remain between the balls.

Class 9 Maths Chapter 14 Very Short Answer Questions

7. Write the formulas for the total surface area and volume of a cuboid of length l, width w and height h.

Answer:

$$\text{TSA}=2(lw+wh+hl)$$

$$V=lwh$$

8. A cylindrical container is open at the top and closed at the bottom. Write the formula for its surface area, ignoring its thickness.

Answer:

Its surface area includes the curved surface and one circular base.

$$\text{Surface area}=2\pi rh+\pi r^2$$

9. State the relationship between the radius r, height h and slant height l of a right circular cone.

Answer:

$$l^2=h^2+r^2$$

10. Write the formula for the volume of a pyramid.

Answer:

$$V=\dfrac{1}{3}\times\text{Area of base}\times\text{Perpendicular height}$$

11. Why are surface area and volume expressed in different units?

Answer:

Surface area measures a two-dimensional surface, so it uses square units such as cm². Volume measures three-dimensional space, so it uses cubic units such as cm³.

Class 9 Maths Chapter 14 Short Answer Questions

12. How many small cubes of side 20 cm can be packed tightly into a cubical box of side 2 m?

Solution:

Convert the side of the box into centimetres:

2 m = 200 cm.

Number of small cubes along each edge:

$$n=\dfrac{200}{20}=10$$

Total number of cubes:

$$N=10\times10\times10=1000$$

Answer: 1,000 cubes.

13. Two cubes, each of volume 125 cm³, are joined face to face. Find the total surface area of the resulting cuboid.

Solution:

Let the side of each cube be a.

$$a^3=125$$

Therefore, a = 5 cm.

The resulting cuboid has dimensions 10 cm × 5 cm × 5 cm.

$$\text{TSA}=2(lw+wh+hl)$$

$$\text{TSA}=2[(10\times5)+(5\times5)+(5\times10)]$$

$$\text{TSA}=2(50+25+50)=250\text{ cm}^2$$

Answer: 250 cm².

14. The radius of cylinder B is twice that of cylinder A, while its height is half that of cylinder A. Find the ratios of their curved surface areas and volumes, A : B.

Solution:

Let cylinder A have radius r and height h.

Cylinder B has radius 2r and height $$\dfrac{h}{2}$$.

Curved surface areas:

$$\text{CSA}_A=2\pi rh$$

$$\text{CSA}_B=2\pi(2r)\left(\dfrac{h}{2}\right)=2\pi rh$$

Therefore, the ratio of curved surface areas is 1 : 1.

Volumes:

$$V_A=\pi r^2h$$

$$V_B=\pi(2r)^2\left(\dfrac{h}{2}\right)=2\pi r^2h$$

Therefore, the ratio of volumes is 1 : 2.

15. The curved surface area of a cone is 308 cm², and its slant height is 14 cm. Find its base radius and total surface area.

Solution:

$$\text{CSA}=\pi rl$$

$$308=\dfrac{22}{7}\times r\times14$$

$$308=44r$$

Therefore, r = 7 cm.

$$\text{TSA}=\pi rl+\pi r^2$$

$$\text{TSA}=308+\dfrac{22}{7}\times7^2$$

$$\text{TSA}=308+154=462\text{ cm}^2$$

Answer: Radius = 7 cm; total surface area = 462 cm².

16. The areas of three faces of a cuboid meeting at one corner are 6 cm², 15 cm² and 10 cm². Find its volume.

Solution:

Let its dimensions be l, w and h.

$$lw=6,\quad wh=15,\quad hl=10$$

Multiplying:

$$(lw)(wh)(hl)=6\times15\times10$$

$$(lwh)^2=900$$

Since the dimensions are positive:

$$V=lwh=\sqrt{900}=30\text{ cm}^3$$

Answer: 30 cm³.

Class 9 Maths Chapter 14 Long Answer Questions

17. A joker’s cap is shaped like a right circular cone with base radius 7 cm and height 24 cm. Find the area of sheet required to make 10 such caps, ignoring wastage.

Solution:

Given radius r = 7 cm and height h = 24 cm.

First, find the slant height:

$$l=\sqrt{h^2+r^2}$$

$$l=\sqrt{24^2+7^2}=\sqrt{576+49}=\sqrt{625}=25\text{ cm}$$

A cap is open at the base, so only its curved surface requires material.

$$\text{Area for one cap}=\pi rl$$

$$\text{Area for one cap}=\dfrac{22}{7}\times7\times25=550\text{ cm}^2$$

For 10 caps:

$$\text{Total area}=10\times550=5500\text{ cm}^2$$

Answer: 5,500 cm² of sheet.

18. A school serves milk to 1,600 students in cylindrical glasses of diameter 7 cm. Each glass contains milk up to a height of 12 cm. How many litres of milk are required?

Solution:

Radius of each glass:

$$r=\dfrac{7}{2}=3.5\text{ cm}$$

Height of milk = 12 cm.

Volume of milk in one glass:

$$V=\pi r^2h$$

$$V=\dfrac{22}{7}\times\dfrac{7}{2}\times\dfrac{7}{2}\times12=462\text{ cm}^3$$

For 1,600 students:

$$\text{Total volume}=462\times1600=739200\text{ cm}^3$$

Since 1,000 cm³ = 1 litre:

$$\text{Milk required}=\dfrac{739200}{1000}=739.2\text{ litres}$$

Answer: 739.2 litres.

19. Show that the volume of a sphere is two-thirds of the volume of the smallest right circular cylinder enclosing it.

Solution:

Let the sphere have radius r.

The smallest enclosing cylinder has:

Radius = r.

Height = 2r.

Volume of the sphere:

$$V_s=\dfrac{4}{3}\pi r^3$$

Volume of the cylinder:

$$V_c=\pi r^2(2r)=2\pi r^3$$

Therefore:

$$\dfrac{V_s}{V_c}=\dfrac{\dfrac{4}{3}\pi r^3}{2\pi r^3}=\dfrac{2}{3}$$

Hence:

$$V_s=\dfrac{2}{3}V_c$$

Answer: The sphere occupies two-thirds of the enclosing cylinder’s volume.

20. A cylindrical glass of radius 4 cm contains water up to a height of 16 cm. How many spherical marbles of radius 1 cm must be dropped into it to raise the water level to 20 cm? Assume the marbles are fully submerged and no water spills.

Solution:

Required rise in water level:

20 − 16 = 4 cm.

Volume of water displaced:

$$V_d=\pi r^2h=\pi\times4^2\times4=64\pi\text{ cm}^3$$

Volume of one marble:

$$V_m=\dfrac{4}{3}\pi(1)^3=\dfrac{4}{3}\pi\text{ cm}^3$$

Number of marbles:

$$N=\dfrac{V_d}{V_m}=\dfrac{64\pi}{\dfrac{4}{3}\pi}=48$$

Answer: 48 marbles.

Common Mistakes to Avoid in Class 9 Maths Chapter 14 Questions

Small errors can change your surface area and volume answers. Check these details while solving:

• Radius and Diameter: Halve the diameter before using a formula that requires the radius.

• Cone Measurements: Use slant height for curved surface area and perpendicular height for volume.

• Hidden Faces: Exclude joined faces when finding the exposed surface area of combined solids.

• Unit Conversion: Remember that 1 m² = 10,000 cm², while 1 m³ = 1,000,000 cm³.

• Rounding: Keep π in your working until the final calculation where possible.

Learn Step by Step with Vedantu’s Class 9 Maths Chapter 14 Important Questions PDF

Vedantu’s Class 9 Maths Chapter 14 important questions with solutions help students follow the reasoning behind each calculation in Math of Space: Surface Area and Volume.

• Worked Methods: See how dimensions, formulas and unit conversions fit together.

• Concept-Based Questions: Check your understanding of solids, cross-sections and volume relationships.

• Answer Checking: Compare your working with the solutions to locate calculation errors.

• Convenient Access: Download the Class 9 Maths Chapter 14 important questions PDF for offline study at your own pace.

Important Study Materials for Class 9 Maths (Ganita Manjari II) Chapter 14 Math of Space: Surface Area and Volume

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