CBSE Important Questions Class 10 Maths Ch12 Surface Areas and Volumes 2025-26

The chapter Surface Areas and Volumes is one of the most practical and concept-based topics in Class 10 Mathematics Syllabus. It connects geometry with real-life applications by teaching you how to calculate areas and volumes of three-dimensional shapes such as cubes, cuboids, cylinders, cones, spheres, and combinations of these solids.

Understanding this chapter helps you develop visualisation and problem-solving skills that are useful not only in mathematics but also in physics, architecture, and engineering.The Surface Areas and Volumes Important Questions for Class 10 Mathematics help you:

• Revise all standard formulas for surface area and volume.
• Strengthen problem-solving using real-life examples.
• Gain speed and accuracy in numerical applications for exams

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IMPORTANT QUESTIONS CLASS 10

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Ch12 Surface Area and Volume Important Questions Class 10 Maths

1. Anishka melted 11 chocolate cubes in a cylindrical cup as shown.

(Note: The figures are not to scale.)

If the length of the side of each cube is k cm and the radius of the cup is r cm, which of these represents the height of the melted chocolate in the cup?

(Note: Take π = 22/7)

a. (7k³/4r) cm

b. (7k³/2r²) cm

c. (7k²/4r) cm

d. (7k²/2r²) cm

Soln. (b) (7k³/2r²) cm

Explanation:

Length of side of each cube= k cm

Radius of the cup= r cm

Number of chocolates melted in the cylindrical cup= 11

We know,

Volume of cube= side³

Volume of cylinder= πr²h

Therefore, we get:

Volume of one chocolate cube= k³

Volume of 11 chocolate cubes= 11k³  ...(i)

Volume of cylindrical cup

= πr²h

= 22r²h/7 ...(ii)

Equating i and ii, we get:

11k³= 22r²h/7

⇒h= 11k³×7/22r²

⇒h= 7k³/2r²

Hence, we get the required answer.

2. A container with a grey hemispherical lid has radius R cm. In figure 1, it contains water upto a height of R cm. It is then inverted as shown in figure 2.

What is the height of water in figure 2?

a. R cm

b. (5R/3) cm

c. 2R cm

d. (7R/3) cm

Soln.(b) (5R/3) cm

Explanation:

We know,

Volume of hemisphere= 2/3πr³

Volume of cylinder= πr²h

Let R be the radius of hemispherical lid and height of water and h be the height of cylinder.

Therefore, we get:

Volume of given cylinder= πR²h cm³

Volume of the hemispherical lid= 2/3πR³

Volume of water inside= πR²×R (given height of water is R)

Therefore, we get:

πR³= πR²h-2/3πR³

πR³= πR²h-2/3R

R= h-2/3R

h= R+2/3R

= 5R/3

Hence, the height of water when it is inverted is 5R/3 cm.

3. A paint roller is a paint application tool used for painting large flat surfaces rapidly and efficiently. One such roller is shown below, which is 26 cm long with an outer diameter of 7 cm.

(Note: The figure is not to scale.)

Find the maximum area of the surface that gets painted when the roller makes 6 complete rotations vertically. Show your work.

(Note: Take π = 22/7)

Soln. Given,

A cylindrical shaped roller with height= 26 cm and diameter= 7 cm.

We know,

Curved surface area of cylinder= 2πrh

Therefore, we get:

CSA of the paint roller= 2×22/7×72×26

= 22×26

= 572 cm²

i.e., Area covered after one rotation= 572 cm²

Area covered after 6 rotations= 6×572

= 3432 cm²

Hence, the required answer is 3432 cm².

4. Shown below is a triangular prism and a cylinder.

On filling the cylinder with water and completely submerging the prism inside the filled cylinder, some water was forced to overflow out of the cylinder. When the prism was removed again, it was noticed that the cylinder had exactly of its water remaining.

If the volume of the cylinder is K litres, what is the volume of the prism? Show your steps and give valid reason.

Soln. Given,

Volume of the cylinder= K

l= Volume of water in it initially

Volume of water remaining in the cylinder= 3/5K l

From the question, we know:

Volume of triangular prism= Volume of water forced to overflow out of the cylinder

Let the volume of prism/ water overflown be x.

Then, Volume of water in the cylinder initially, 

K= Volume of water overflown+Volume of water remaining in the cylinder

Therefore, we get:

Volume of water overflown/ prism

= K-3/5K

= 5K-3K/5

= 2K/5 l

Hence, we get the volume of prism is 2/5K l.

5. Shown below is a cake that Subodh is baking for his brother's birthday. The cake is 21 cm tall and has a radius of 15 cm. He wants to surprise his brother by filling gems inside the cake. In order to do that, he removes a cylindrical portion of cake out of the centre as shown. The piece that is removed is 21 cm tall.

If the cake weighs 0.5 g per cubic cm and the weight of the cake that is left after removing the central portion is 6600 g, find the radius of the central portion that is cut. Show your steps.

(Note: Take π = 22/7)

Soln. Given,

Height of the cylindrical cake= 21 cm= height of the piece that is removed

Radius= 15 cm

We know,

Volume of cylinder= πr²h

Therefore, we get:

Volume of cake

= 227×15×15×21

= 14850 cm³

Weight of cake per 1 cm³

= 0.5 g

Then, weight of cake per 14850 cm³

= 14850×0.5

= 7425 g

Given that the weight of the cake after removing the middle portion= 6600 g

Thus, the weight of the portion removed

= 7425-6600

= 825 g

The volume of the portion removed

= 8250.5

= 1650 cm³

Therefore, we get:

Volume of the portion removed, 1650 cm³

= πr²×21

⇒r²= (1650×7)/(21×22)

⇒r²= 25

= 5 cm

Hence, the radius of the portion removed is 5 cm.

6. Dinesh is building a greenhouse in his farm as shown below. The base of the greenhouse is circular having a diameter of 12 m and it has a hemispherical dome on top.

(Note: The image is not to scale.)

How much will it cost him to cover the walls and top of the greenhouse with transparent plastic, if the plastic sheet costs Rs 77 per sq m? Show your steps. 

(Note: Take π = 22/7)

Soln. Given,

Diameter of hemisphere= 12 m

Height of the cylindrical wall= 2 m

We know,

Curved surface area of hemisphere= 2πr²

Curved surface area of cylinder= 2πrh

Therefore, we get:

CSA of top of the greenhouse= 2×π×12/2×12/2

= 72 π m²

CSA of wall= 2×π×12/2×2= 24π m2

Total surface area to be covered with plastic= 72π+24π= 96 π m2

Cost of plastic to cover the whole greenhouse= ₹77×96π= ₹23232

Hence, the required answer is ₹23232.

7. Shifali made a lampshade using cane web as shown below.

Find the minimum number of sheets of cane web required to make this lamp if each sheet has an area of 44 square inches.

(Note: Take π = 22/7)

Soln. We know, 

Curved surface area of cylinder= 2πrh

Curved surface area of frustum of cone= πLr+r'

Therefore, we get:

CSA of cylindrical part

= 2×π×16/2×5

= 80π inch²

CSA of frustum of cone

= π×5×5+8

= 65π inch²

Total area of the lampshade

= 80π+65π

= 145π inch²

Area of cane web sheet= 44 inch²

Minimum number of cane web sheets required= 145π/44

= (145×22)/(44×7)

= 10.357≈10.4

That is, we require minimum of 11 sheets to make this lamp as we will get the sheet in whole numbers.

Hence, the required answer is 11.

8. Shown below is a solid made of a cone, a cylinder and a hemisphere.

(Note: The figure is not to scale.)

Prove that the total volume of the solid is twice the volume of the cylinder.

Soln.

We know,

Volume of cylinder= πr²h

Volume of cone= πr²h³

Volume of hemisphere= 2/3πr³

Therefore, we get:

Volume of cylinder= πk²×k

= πk³cm³

Volume of cone

= π×k²×k/3

= πk³/3 cm³

Volume of hemisphere= 2/3π×k³

^‍= 2πk³/3 cm³

Total volume of solid= π(k³+k³/3+2k/3)/3 cm³

= π3k³+k³+2k³/3

= π6k³3= 2πk³ cm³ ...(i)

Volume of cylinder= πk³ cm³ ...(ii)

Comparing both (i) and (ii), we get :

The volume of Solid= 2×volume of cylinder

Hence, proved.

9. A cloche is used to cover dishes before serving. Shown below is a hemispherical glass cloche of radius 13 cm. Kanan wants to use it to cover a cylindrical cake of volume 3168 cm³.

Find one set of values of radius and height of the cake, such that the cloche does not touch the cake when covered. Show your steps.

(Note: Take π = 22/7)

Soln.

Given,

Volume of cylindrical cake= 3168 cm³

Radius of hemispherical cloche= 13 cm

We know, 

Volume of cylinder= πr²h

Volume of hemisphere= 2/3πr³

Therefore, we get:

Volume of the cake, 3168= π×r²h

⇒r²h= 3168×7/22

= 1008

According to the question, we know:

r and h of the cylindrical cake should be less than 13 cm, the radius of cloche.

Thus, prime factorising 1008, we have:

1008= 2×2×2×2×3×3×7

Pairing the factors, 1008= 2×2×2×2×3×3×7.

That is, 1008= 2×2×32×7= 122×7

⇒r²h= 12²×7

⇒r= 12 cm and h= 7 cm

Hence, the radius and height of the cake are 12 cm and 7 cm respectively.

10. The surface area of a solid spherical ball is S cm2. It is cut into 4 identical pieces, as shown below.

Find the total surface area of 4 identical pieces of the solid spherical ball in terms of S. Show your work.

Soln. Let r be the radius of the spherical ball.

Therefore, we get:

The surface area of sphere, s= 4πr² cm² ...(i)

Surface area of one of the identical pieces= 144πr²+π^r2

= 2πr² cm²

Total surface area of 4 identical pieces= 4X2πr²

= 8πr² cm² ...(ii)

Substituting the equation (i) in (ii), we get:

Total surface area of 4 identical pieces= 2×4πr²= 2s cm²

Hence, the required answer is 2s cm².

Some More Important Question Answers of Class 10 Maths Surface Areas and Volumes

Q1. Find the total surface area and volume of a cube of 8 cm.

Ans. Side = 8 cm
TSA = 6a² = 6 × 8² = 6 × 64 = 384 cm²
Volume = a³ = 8³ = 512 cm³

Answer: TSA = 384 cm², Volume = 512 cm³

Q2. Find the surface area and volume of a cuboid of length 12 cm, breadth 8 cm, and height 5 cm.

Ans. TSA = 2(lb + bh + hl)
= 2(12×8 + 8×5 + 5×12)
= 2(96 + 40 + 60)
= 2×196 = 392 cm²

Volume = l × b × h = 12×8×5 = 480 cm³

Answer: TSA = 392 cm², Volume = 480 cm³

Q3. A cylinder has radius 7 cm and height 15 cm. Find its curved surface area, total surface area, and volume.

Ans. CSA = 2πrh = 2×(22/7)×7×15 = 660 cm²
TSA = 2πr(h + r) = 2×(22/7)×7×(15+7) = 968 cm²
Volume = πr²h = (22/7)×7×7×15 = 2310 cm³

Answer: CSA = 660 cm², TSA = 968 cm², Volume = 2310 cm³

Q4. Find the curved surface area and volume of a cone of radius 7 cm and slant height 25 cm.

Ans. CSA = πrl = (22/7)×7×25 = 550 cm²
Volume = (1/3)πr²h
To find h, use Pythagoras: h² = l² – r² = 25² – 7² = 625 – 49 = 576 → h = 24 cm
So, Volume = (1/3)×(22/7)×7×7×24 = 1232 cm³

Answer: CSA = 550 cm², Volume = 1232 cm³

Q5. Find the total surface area of a sphere of radius 7 cm.

Ans. TSA = 4πr² = 4×(22/7)×7×7 = 616 cm²

Answer: 616 cm²

Q6. Find the volume of a sphere of radius 3.5 cm.

Ans. Volume = (4/3)πr³
= (4/3)×(22/7)×3.5×3.5×3.5
= (4/3)×(22/7)×42.875
= 219.8 cm³

Answer: 219.8 cm³

Q7. Find the total surface area and volume of a hemisphere of radius 7 cm.

Ans. TSA = 3πr² = 3×(22/7)×7×7 = 462 cm²
Volume = (2/3)πr³ = (2/3)×(22/7)×7×7×7 = 718.7 cm³

Answer: TSA = 462 cm², Volume = 718.7 cm³

Q8. Find the curved surface area of a cone of slant height 20 cm and base diameter 14 cm.

Ans. Radius = 14/2 = 7 cm
CSA = πrl = (22/7)×7×20 = 440 cm²

Answer: 440 cm²

Q9. Find the total surface area and volume of a cylinder of height 10 cm and diameter 14 cm.

Ans. Radius = 7 cm
TSA = 2πr(h + r) = 2×(22/7)×7×(10+7) = 748 cm²
Volume = πr²h = (22/7)×7×7×10 = 1540 cm³

Answer: TSA = 748 cm², Volume = 1540 cm³

Q10. A solid metallic cylinder of radius 3 cm and height 5 cm is melted to form a sphere. Find the radius of the sphere.

Ans. Volume of cylinder = πr²h = (22/7)×3×3×5 = 141.4 cm³
Let sphere radius = R
Volume of sphere = (4/3)πR³ = 141.4
(4/3)πR³ = 141.4 → R³ = (141.4×3)/(4×3.14) = 33.8 → R = 3.2 cm

Answer: Radius of sphere = 3.2 cm

Q11. Find the volume of a cone whose radius is 7 cm and height 24 cm.

Ans. Volume = (1/3)πr²h = (1/3)×(22/7)×7×7×24 = 1232 cm³

Answer: 1232 cm³

Q12. Find the height of a cone whose slant height is 10 cm and base radius is 6 cm.

Ans. h² = l² – r² = 10² – 6² = 100 – 36 = 64
h = 8 cm

Answer: Height = 8 cm

Q13. The height and radius of a cylinder are in the ratio 3:1. If its volume is 924 cm³, find its height and radius.

Ans. Let radius = r, height = 3r
Volume = πr²h = 924
(22/7)×r²×3r = 924 → (66/7)r³ = 924 → r³ = (924×7)/66 = 98 → r = 4.6 cm
Height = 3r = 13.8 cm

Answer: r = 4.6 cm, h = 13.8 cm

Q14. A cone and a cylinder have the same base radius and height. Find the ratio of their volumes.

Ans. Volume of cone = (1/3)πr²h
Volume of cylinder = πr²h
Ratio = 1:3

Answer: 1:3

Q15. Find the volume of a hemisphere of radius 3.5 cm.

Ans. Volume = (2/3)πr³ = (2/3)×(22/7)×3.5×3.5×3.5 = 89.8 cm³

Answer: 89.8 cm³

Q16. Find the area of the curved surface of a hemisphere of radius 7 cm.

Ans. CSA = 2πr² = 2×(22/7)×7×7 = 308 cm²

Answer: 308 cm²

Q17. A cylinder and a cone have the same base radius 7 cm and height 14 cm. Find the difference in their volumes.

Ans. Volume of cylinder = πr²h = (22/7)×7×7×14 = 2156 cm³
Volume of cone = (1/3)πr²h = (1/3)×(22/7)×7×7×14 = 718.7 cm³
Difference = 2156 – 718.7 = 1437.3 cm³

Answer: 1437.3 cm³

Q18. The diameter of a sphere is 14 cm. Find its surface area and volume.

Ans. Radius = 7 cm
Surface area = 4πr² = 4×(22/7)×7×7 = 616 cm²
Volume = (4/3)πr³ = (4/3)×(22/7)×7×7×7 = 1437.3 cm³

Answer: SA = 616 cm², Volume = 1437.3 cm³

Q19. A cone, a hemisphere and a cylinder have equal bases and the same height is 7 cm. Find the ratio of their volumes.

Ans. Let radius = r
Volume of cone = (1/3)πr²h = (1/3)πr²×7 = (7πr²)/3
Volume of hemisphere = (2/3)πr³
Volume of cylinder = πr²h = 7πr²

Ratio = (7πr²/3) : (2πr³/3) : (7πr²)
Simplify → 7 : 2r : 21 → depends on r, for r = 7 cm → 7 : 14 : 21 → 1 : 2 : 3

Answer: Ratio = 1 : 2 : 3

Q20. A metallic sphere of radius 4.2 cm is melted and recast into smaller spheres of radius 0.7 cm. Find how many such small spheres can be made.

Ans. Volume of big sphere = (4/3)π(4.2)³ = (4/3)×3.14×74.1 = 310.3 cm³
Volume of one small sphere = (4/3)π(0.7)³ = (4/3)×3.14×0.343 = 1.436 cm³
Number = 310.3 ÷ 1.436 ≈ 216

Answer: 216 small spheres

Extra Questions for Practice!

Q1. Find the total surface area of a cube of 9 cm.

Q2. Find the volume of a cylinder of radius 3.5 cm and height 10 cm.

Q3. Find the curved surface area of a cone of slant height 20 cm and radius 12 cm.

Q4. Find the volume of a hemisphere whose diameter is 14 cm.

Q5. A solid sphere of radius 5 cm is melted into small cones of height 5 cm and radius 2 cm. Find the number of cones.

Q6. Find the height of a cone whose volume is 308 cm³ and radius is 7 cm.

Q7. The total surface area of a sphere is 616 cm². Find its radius.

Q8. A right circular cylinder just encloses a sphere of radius 7 cm. Find the height of the cylinder.

Q9. Find the curved surface area of a frustum of a cone having radii 7 cm, 3.5 cm and slant height 10 cm.

Q10. A conical tent is 10 m high and base radius 14 m. Find the area of canvas required to make it.

How to Ace These Class 10 Surface Areas and Volumes Important Questions

This chapter can be mastered easily through understanding and repeated practice. Follow this step-by-step approach to prepare effectively:

Step 1: Revise Basic Formulas

Memorise the surface area and volume formulas of all 3D shapes from the NCERT textbook:

• Cube: Surface Area = 6a², Volume = a³
• Cuboid: Surface Area = 2(lb + bh + hl), Volume = l × b × h
• Cylinder: Curved Surface Area (CSA) = 2πrh, Total Surface Area (TSA) = 2πr(h + r), Volume = πr²h
• Cone: CSA = πrl, TSA = πr(l + r), Volume = ⅓πr²h
• Sphere: Surface Area = 4πr², Volume = ⁴⁄₃πr³

Step 2: Understand Derivations Conceptually

Knowing how formulas are derived from 2D shapes (like circles forming cylinders or cones) helps in remembering them longer and applying them confidently.

Step 3: Practise Combination Solids

Many exam questions involve combinations such as a cone on a hemisphere or a cylinder topped with a cone. Learn to split such solids into parts and apply respective formulas step by step.

Step 4: Learn Unit Conversions

Always convert all given measurements into the same units before solving (for example, cm to m). This avoids calculation mistakes and ensures correct answers.

Step 5: Revise Frustum of a Cone

Understand formulas for surface area and volume of a frustum, as it’s frequently asked in 3–5 mark questions.

Step 6: Apply Concepts in Real-Life Problems

Practise word problems related to tanks, pipes, ice-cream cones, hemispherical bowls, etc. These strengthen your conceptual understanding and accuracy.

Why should you use Class 10 Ch12 extra questions?

Practicing more than just the textbook questions is key to success in this chapter. Here's why practicing surface area and volume class 10 extra questions is important: 

• NCERT Textbooks often cover the basic concepts, but real exams challenge you with diverse and tricky problems. By solving extra questions, you check your preparation by solving different types of problems, including those involving complex shapes or real-life applications.
• Identifying weak areas is one of the most important aspects of studying mathematics. When you attempt a wide variety of questions, it becomes easier to spot which type of problems you struggle with. 
• In exams, speed and accuracy matter. Solving a large number of surface area and volume class 10 important questions helps you improve both your speed and precision. More practice makes you faster at applying the correct formula and solving the problem in time.

Frequently Asked Questions

Q1. How many marks are allotted to this chapter in Class 10 board exams?

Ans. This chapter generally carries 7 to 9 marks, including direct numerical and application-based questions.

Q2. What types of questions are asked in exams?

Ans. You can expect formula-based questions, application questions involving combination solids, and problems related to conversion of one solid into another.

Q3. How can I avoid mistakes in surface area and volume problems?

Ans. Write formulas clearly before substitution, check units, and ensure correct identification of radius, height, and slant height.

Q4. Which topics are most important?

Ans. Cylinders, cones, spheres, and combination solids are the most frequently asked topics.

Q5. How can I quickly revise this chapter before exams?

Ans. Revise all formulas from NCERT, practise 5–6 application questions, and review one combination solid problem.