What is a frustum of a cone?

A frustum is the solid shape formed when a cone is cut by a plane parallel to its base, removing the top portion. It looks like a bucket or lampshade with two circular ends of different radii. Its volume is (1/3) pi h (R squared + r squared + Rr).

How do you find the surface area of a combined solid?

Add the curved surface areas of individual parts but exclude the areas where shapes join. For example, a toy made of a cone on a hemisphere: TSA = CSA of cone + CSA of hemisphere. The flat circular faces where they meet are not included.

What stays the same when a solid is melted and recast?

The volume remains the same. You equate the volume of the original solid to the volume of the new solid(s) to find unknown dimensions. Surface area generally changes.

How do you convert between cm cubed and litres?

1 litre = 1000 cm cubed. To convert from cm cubed to litres, divide by 1000. This conversion is commonly needed in problems asking for the capacity of containers.

How many marks does Chapter 12 carry in CBSE board exams?

Chapter 12 typically carries 5 to 8 marks. Popular questions include finding the surface area or volume of combined solids (4-5 marks), conversion problems (3-4 marks), and frustum problems (3-4 marks).

What is the slant height formula for a frustum?

The slant height of a frustum is l = square root of (h squared + (R - r) squared), where h is the height and R, r are the two radii. Note: it uses the difference of radii, not the radii themselves.

How do you find the volume of a toy that is a cone mounted on a hemisphere?

Volume = Volume of cone + Volume of hemisphere = (1/3) pi r squared h + (2/3) pi r cubed. You can factor this as (1/3) pi r squared (h + 2r). Make sure h is the height of the cone only, not the total height.

What is a frustum of a cone?

A frustum is the solid shape formed when a cone is cut by a plane parallel to its base, removing the top portion. It looks like a bucket or lampshade with two circular ends of different radii. Its volume is (1/3) pi h (R squared + r squared + Rr).

How do you find the surface area of a combined solid?

Add the curved surface areas of individual parts but exclude the areas where shapes join. For example, a toy made of a cone on a hemisphere: TSA = CSA of cone + CSA of hemisphere. The flat circular faces where they meet are not included.

What stays the same when a solid is melted and recast?

The volume remains the same. You equate the volume of the original solid to the volume of the new solid(s) to find unknown dimensions. Surface area generally changes.

How do you convert between cm cubed and litres?

1 litre = 1000 cm cubed. To convert from cm cubed to litres, divide by 1000. This conversion is commonly needed in problems asking for the capacity of containers.

How many marks does Chapter 12 carry in CBSE board exams?

Chapter 12 typically carries 5 to 8 marks. Popular questions include finding the surface area or volume of combined solids (4-5 marks), conversion problems (3-4 marks), and frustum problems (3-4 marks).

What is the slant height formula for a frustum?

The slant height of a frustum is l = square root of (h squared + (R - r) squared), where h is the height and R, r are the two radii. Note: it uses the difference of radii, not the radii themselves.

How do you find the volume of a toy that is a cone mounted on a hemisphere?

Volume = Volume of cone + Volume of hemisphere = (1/3) pi r squared h + (2/3) pi r cubed. You can factor this as (1/3) pi r squared (h + 2r). Make sure h is the height of the cone only, not the total height.

Solved Examples

NCERT Solutions for Class 10 Maths Chapter 12: Surface Areas and Volumes — Free PDF

Step-by-step solutions for combined solids, conversion of solids, and frustum of a cone problems.

CBSEClass 10

The SparkEd Authors (IITian & Googler)15 March 202635 min read

Chapter Overview: Surface Areas and Volumes

$Chapter 12 is one of the most application-oriented chapters in Class 10 Maths. It extends your Class 9 knowledge of surface areas and volumes to handle three new types of problems: combined solids (shapes made by joining basic solids like cylinders, cones, and hemispheres), conversion of solids (melting and recasting one solid into another), and the frustum of a cone (a cone with its top cut off).$

$\pi$

$The chapter has 5 exercises covering: - Exercise 12.1: Surface area of combined solids - Exercise 12.2: Volume of combined solids - Exercise 12.3: Conversion of solids (same volume, different shape) - Exercise 12.4: Frustum of a cone - Exercise 12.5: Mixed problems$

$This chapter typically carries 5-8 marks in the CBSE board exam.$

Key Concepts and Formulas

$Master these formulas before attempting the exercises.$

Basic Solid Formulas

$Solid$	$CSA/LSA$	$TSA$	$Volume$
$Cylinder$	$2\pi rh$	$2\pi r(r+h)$	$\pi r^2 h$
$Cone$	$\pi rl$	$\pi r(r+l)$	$\dfrac{1}{3}\pi r^2 h$
$Sphere$	$4\pi r^2$	$4\pi r^2$	$\dfrac{4}{3}\pi r^3$
$Hemisphere$	$2\pi r^2$	$3\pi r^2$	$\dfrac{2}{3}\pi r^3$

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

Frustum of a Cone

$R$

l = \sqrt{h^2 + (R-r)^2}

\text{CSA} = \pi(R + r)l

\text{TSA} = \pi(R+r)l + \pi R^2 + \pi r^2

\text{Volume} = \dfrac{1}{3}\pi h(R^2 + r^2 + Rr)

Key Principles

$Combining solids: When two solids are joined, the TSA of the combined solid equals the sum of the CSAs of each part, excluding the areas where they are joined.$

$Converting solids: When a solid is melted and recast into a different shape, the volume remains the same . Equate volumes to find unknown dimensions.$

$1$

Exercise 12.1 — Surface Area of Combined Solids (Solved)

$Combined solids are formed by joining basic shapes. The key is to identify which surfaces are exposed.$

Problem 1: Cone on a Hemisphere

$= 3.5$

$r = 3.5$

$l = \sqrt{12^2 + 3.5^2} = \sqrt{144 + 12.25} = \sqrt{156.25} = 12.5$

$TSA = CSA of cone + CSA of hemisphere:$

= \pi r l + 2\pi r^2 = \pi r(l + 2r)

= \dfrac{22}{7} \times 3.5 \times (12.5 + 7) = \dfrac{22}{7} \times 3.5 \times 19.5 = 11 \times 19.5 = 214.5 \text{ cm}^2

$= 214.5$

Problem 2: Capsule (Cylinder + Two Hemispheres)

$= 14$

$r = 2.5$

$\times$

= 2\pi r \times 9 + 2 \times 2\pi r^2 = 2\pi r(9 + 2r)

= 2 \times \dfrac{22}{7} \times 2.5 \times (9 + 5) = 2 \times \dfrac{22}{7} \times 2.5 \times 14 = 220 \text{ mm}^2

$= 220$

Problem 3: Tent (Cylinder + Cone)

$^2$

$= 2\pi r h = 2 \times \dfrac{22}{7} \times 10.5 \times 3 = 198$

$= \pi r l = \dfrac{22}{7} \times 10.5 \times 21 = 693$

$= 198 + 693 = 891$

$= 891 \times 2 =$

$=$

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Exercise 12.3 — Conversion of Solids (Solved)

$When solids are melted and recast, the volume is conserved.$

Problem 1: Sphere Recast into Cylinder

$Problem: A metallic sphere of radius 4.2 cm is melted and recast into a cylinder of radius 6 cm. Find the height.$

$Solution: Volume of sphere = Volume of cylinder:$

\dfrac{4}{3}\pi (4.2)^3 = \pi (6)^2 \times h

\dfrac{4}{3} \times 74.088 = 36h

h = \dfrac{4 \times 74.088}{3 \times 36} = \dfrac{296.352}{108} = 2.744 \text{ cm}

$= 2.74$

Problem 2: Cuboid into Coins

$5.5 \times 10 \times 3.5$

$= 5.5 \times 10 \times 3.5 = 192.5$

$= \pi r^2 h = \dfrac{22}{7} \times (0.875)^2 \times 0.2 = 0.48125$

$= \dfrac{192.5}{0.48125} = 400$

$400$

Problem 3: Cone into Spheres

$Problem: A cone of height 24 cm and radius 6 cm is melted to form spheres of radius 2 cm each. How many spheres are formed?$

$= \dfrac{1}{3}\pi (6)^2 \times 24 = \dfrac{1}{3}\pi \times 864 = 288\pi$

$= \dfrac{4}{3}\pi (2)^3 = \dfrac{32\pi}{3}$

$= \dfrac{288\pi}{\frac{32\pi}{3}} = \dfrac{288 \times 3}{32} = \dfrac{864}{32} = 27$

$27$

Exercise 12.4 — Frustum of a Cone (Solved)

$R$

Problem: Bucket-shaped Frustum

$= 30$

$R = 15$

$Capacity (Volume):$

V = \dfrac{1}{3}\pi h(R^2 + r^2 + Rr) = \dfrac{1}{3} \times \dfrac{22}{7} \times 24 \times (225 + 100 + 150)

= \dfrac{22 \times 8 \times 475}{7} = \dfrac{83600}{7} = 11942.86 \text{ cm}^3

$\approx 11.94$

$l = \sqrt{24^2 + (15-10)^2} = \sqrt{576 + 25} = \sqrt{601} \approx 24.52$

$= \pi(R+r)l + \pi r^2$

= \dfrac{22}{7}(25)(24.52) + \dfrac{22}{7}(100) \approx 1926.57 + 314.29 \approx 2240.86 \text{ cm}^2

$\approx 2240.86 \times 1.50 \approx$

Problem: Frustum Volume and Surface Area

$Problem: A frustum has top radius 5 cm, bottom radius 10 cm, and height 12 cm. Find volume, CSA, and TSA.$

$R = 10$

$l = \sqrt{12^2 + (10-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13$

$= \dfrac{1}{3} \times \dfrac{22}{7} \times 12 \times (100 + 25 + 50) = \dfrac{22 \times 4 \times 175}{7} = \dfrac{15400}{7} = 2200$

$= \pi(R+r)l = \dfrac{22}{7} \times 15 \times 13 = \dfrac{4290}{7} \approx 612.86$

$= 612.86 + \pi(10)^2 + \pi(5)^2 = 612.86 + 314.29 + 78.57 = 1005.72$

Worked Examples — Additional Practice

$More examples covering common exam patterns.$

Example 1: Water in a Cylindrical Vessel with Sphere

$Problem: A sphere of diameter 6 cm is dropped into a cylindrical vessel of diameter 12 cm partly filled with water. Find the rise in water level.$

$Solution: Volume of sphere = Volume of water displaced (cylinder slice).$

\dfrac{4}{3}\pi (3)^3 = \pi (6)^2 \times h

\dfrac{4}{3} \times 27 = 36h

36 = 36h \implies h = 1 \text{ cm}

$1$

Example 2: Wooden Toy (Hemisphere + Cone)

$= 3.5$

$r = 3.5$

$= \dfrac{2}{3}\pi r^3 + \dfrac{1}{3}\pi r^2 h$

Example 3: Rain Water Harvesting

$Problem: Rain water from a flat circular roof of radius 14 m flows into a cylindrical tank of diameter 2 m. If the rainfall is 5 cm, find the rise in water level in the tank.$

$= \pi (14)^2 \times 0.05 = 9.8\pi$

$= \pi (1)^2 \times h = \pi h$

$9.8\pi = \pi h \implies h = 9.8$

$9.8$

Common Mistakes to Avoid

$Mistake 1: Including joined surfaces in TSA. When a cone sits on top of a cylinder, the flat circular face of the cone and the top face of the cylinder are not exposed. Do not include them in the total surface area.$

$Mistake 2: Equating surface areas instead of volumes. When a solid is melted and recast, only the VOLUME stays the same. The surface area generally changes.$

$4$

$1$

$l = \sqrt{h^2 + (R-r)^2}$

Practice Questions with Answers

$Test yourself.$

Q1: Volume of Combined Solid

$= 27$

$= 27 - 14 = 13$

Q2: Number of Spherical Balls

$Question: A cylindrical vessel of radius 6 cm and height 15 cm is full of water. The water is poured into spherical balls of radius 3 cm. How many balls can be made?$

$= \pi(6)^2(15) = 540\pi$

Q3: Frustum CSA

$Question: A frustum has radii 8 cm and 4 cm, and slant height 10 cm. Find the CSA.$

$= \pi(R+r)l = \dfrac{22}{7} \times 12 \times 10 = \dfrac{2640}{7} \approx 377.14$

Exam Tips for Surface Areas and Volumes

$Tip 1 — Identify exposed surfaces in combined solids. Surfaces that are joined together are NOT included in TSA.$

$Tip 2 — For conversion problems, equate volumes, not surface areas. Volume is conserved when a solid is melted.$

$Tip 3 — Draw and label diagrams with all dimensions before starting calculations. This helps you visualise which formula to apply.$

$l = \sqrt{h^2 + (R-r)^2}$

$1$

$\pi$

$\pi = \dfrac{22}{7}$

$Tip 8 — In "how many" problems, the answer must be a whole number. Round down if the exact answer is not an integer.$

$Practise on SparkEd's Surface Areas and Volumes page .$

Key Takeaways

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Frequently Asked Questions

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NCERT Solutions for Class 10 Maths Chapter 12: Surface Areas and Volumes — Free PDF

Chapter Overview: Surface Areas and Volumes

Key Concepts and Formulas

Basic Solid Formulas

Frustum of a Cone

Key Principles

Exercise 12.1 — Surface Area of Combined Solids (Solved)

Problem 1: Cone on a Hemisphere

Problem 2: Capsule (Cylinder + Two Hemispheres)

Problem 3: Tent (Cylinder + Cone)

Exercise 12.3 — Conversion of Solids (Solved)

Problem 1: Sphere Recast into Cylinder

Problem 2: Cuboid into Coins

Problem 3: Cone into Spheres

Exercise 12.4 — Frustum of a Cone (Solved)

Problem: Bucket-shaped Frustum

Problem: Frustum Volume and Surface Area

Worked Examples — Additional Practice

Example 1: Water in a Cylindrical Vessel with Sphere

Example 2: Wooden Toy (Hemisphere + Cone)

Example 3: Rain Water Harvesting

Common Mistakes to Avoid

Practice Questions with Answers

Q1: Volume of Combined Solid

Q2: Number of Spherical Balls

Q3: Frustum CSA

Exam Tips for Surface Areas and Volumes

Key Takeaways

Practice These Topics on SparkEd

Frequently Asked Questions

What is a frustum of a cone?

How do you find the surface area of a combined solid?

What stays the same when a solid is melted and recast?

How do you convert between cm cubed and litres?

How many marks does Chapter 12 carry in CBSE board exams?

What is the slant height formula for a frustum?

How do you find the volume of a toy that is a cone mounted on a hemisphere?

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