CBSE Class 10 Maths Most Important Questions Surface Areas & Volumes with Solutions

Mathematics is often seen as a subject of formulas and numbers, but it teaches us how to solve real-world problems using logical thinking and practical applications. One such important topic in Class 10 mathematics is surface area and volume. This chapter holds immense importance, not only for exams but also for its use in daily life. Whether you are calculating the paint required to cover a wall, measuring the capacity of a water tank, or figuring out the dimensions of packaging materials, concepts of surface area and volume come into play.

Let us see the importance of surface area and volume, focusing on how understanding these concepts equips students to solve real-world problems. Moreover, we will also understand why it’s important to practice surface area and volume class 10 extra questions and class 10 surface area and volume extra questions to excel in this chapter.

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Chapter 12 Surface Area and Volume: Important Questions

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.

What is Surface Area and Volume Class 10?

Before understanding how to approach surface area and volume in class 10 important questions, let’s understand the two important terms:

Surface Area: This refers to the total area that the surface of a 3D object occupies.

For example, when we paint a box, the surface area helps us determine how much paint is needed to cover the entire outside of the box.

The surface area of different shapes like cubes, cuboids, spheres, and cylinders has specific formulas, making it easier to calculate.

Volume: Volume measures the amount of space that a 3D object occupies.

For example, volume is useful when we need to find out how much water can fit inside a container or how much sand is required to fill a pit.

Like surface area, each shape has its own volume formula, whether it’s a cube, cuboid, cylinder, cone, or sphere.

Why is the topic surface area and volume class 10 important?

Surface area and volume aren’t just mathematical concepts; they are deeply rooted in our daily lives. Here's why these concepts are essential:

• These calculations are used frequently, from constructing buildings to filling swimming pools. Calculating surface areas helps determine the amount of material required for construction, painting, or covering surfaces. Volume calculations are important when dealing with storage capacities, whether it’s the fuel in your car or the water in a tank.
• Mastery of this topic forms the basis for advanced mathematics. In higher studies, especially in fields like engineering, architecture, and physics, surface area and volume concepts are extended to more complex shapes and applications.
• Many competitive exams, such as JEE and various state-level engineering exams, include problems related to surface area and volume. These exams expect students to have a solid understanding and the ability to quickly solve related problems.
• Solving questions related to surface area and volume enhances your logical thinking and problem-solving abilities. This chapter isn't just about memorizing formulas but about understanding how to apply them to different geometrical shapes.

Why should you use Class 10 surface area and Volume Class 10 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: 

• 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.

How do you solve important surface area and volume questions? 

Let us see some tips for solving important questions about surface area and volume: 

• When solving surface area or volume questions, the first step is to carefully understand the shape involved. Is it a cube, cuboid, cylinder, or a combination of shapes? Many times, the problem may involve composite figures, such as a hemisphere attached to a cylinder, which require breaking down into simpler shapes.
• Though it’s important to understand the logic behind the formulas, memorizing them is equally important for quick calculations. Ensure that you are familiar with the formulas for the surface area and volume of the following common shapes: Cube, cuboid, sphere, cylinder, and cone
• A common challenge in surface area and volume problems is visualizing the 3D object. If you can’t see the object in your mind, sketching it out on paper can help immensely. This makes it easier to identify the dimensions (like height, radius, or length) and which parts of the shape are relevant for surface area or volume calculations.
• Always ensure that all measurements are in the same units before performing any calculation. Surface area is measured in square units and volume in cubic units. Incorrect unit conversions are a common source of errors in surface area and volume problems.

Mastering the concepts of surface area and volume is essential not just for Class 10 exams but also for practical applications in everyday life. To achieve proficiency in this chapter, students must consistently practice surface area and volume class 10 extra questions and explore a variety of problem types.

By combining a strong understanding of formulas and Class 10 Surface Area and Volume Extra Questions, students can confidently approach even the most challenging problems.