Assertion (A): A toy is in the shape of a right circular cone mounted on a hemisphere. The radius of the hemisphere is 3.5 cm and the total height of the toy is 15.5 cm. The total surface area of the toy is 214.5 cm². Reason (R): The total surface area of the combined solid is the sum of the total surface areas of the individual solids.

Both A and R are true and R is the correct explanation of A.

Both A and R are true but R is not the correct explanation of A.

If the ratio of the volume to the surface area of a sphere is 1 : 3, find the radius of the sphere.

The radius of the sphere is 1 unit.

A well of diameter 4 m is dug 21 m deep. The earth taken out of it is spread evenly all around it to form an embankment of width 2 m. Find the height of the embankment.

The height of the embankment is 7 m.

A solid is made by attaching a hemisphere to the larger base of a frustum of a cone. The hemisphere has radius 9 cm and exactly covers the larger base of the frustum. The smaller base radius of the frustum is 5 cm and the frustum height is 8 cm. Calculate the total surface area of the solid. (Use π = 3.14 and √5 = 2.236)

(187π + 56π√5) cm² or approximately 980.36 cm²

A camping tent is designed as a cylinder of diameter 6 m and height 4 m, surmounted by a cone of the same diameter such that the highest point of the tent is 7 m from the ground. The tent is made of canvas and sits on the ground, with no base canvas. (1) Find the slant height of the conical part. (2) Calculate the area of canvas required to make the tent (excluding the base). (Take π = 3.14) (3) What is the volume of air inside the tent?

Slant height = 3√2 m (≈4.24 m), Canvas area = 115.30 m², Volume = 141.30 m³

Class 10 Maths Chapter 12: Surface Areas and Volumes — Important Questions & Sample Paper

Q: What are the typical problem types in CBSE Class 10 Surface Areas and Volumes?

Exam questions usually fall into three categories: (1) finding surface area/volume of combined solids like toys, tents, or decorative items; (2) conversion of solids (melting and recasting), often with material loss; and (3) frustum of a cone, focusing on curved surface area, total surface area, and volume. Real-life contexts like wells, embankments, and canvas tents are common. Students must carefully note which surfaces are exposed and which are in contact to avoid adding extra areas.

Q: How can students avoid mistakes when calculating surface areas of combined solids?

The key is to visualize the object and identify all visible surfaces. For example, when a cone is placed on a cylinder, the base of the cone and the top base of the cylinder are in contact, so they are not part of the external surface area and should be subtracted. Always sketch the figure and label dimensions. Also, confirm whether the problem asks for curved surface area or total surface area, and whether the base is included.

Q: What is the best way to remember the frustum formulas?

The frustum is essentially a cone with the top removed. Instead of memorizing separate formulas, many students find it easier to derive the volume by subtracting the small cone from the large cone, and the curved surface area as πl(r1+r2), where l is the slant height. For total surface area, add the areas of both circular ends: π(r1^2 + r2^2) + πl(r1+r2). The slant height l can be found using l = √[h² + (r1 - r2)²].

Q: How should one handle recasting problems, especially when some metal is lost?

Start by equating the volume of the original solid(s) to the total volume of the new solids, accounting for any loss. For instance, if 10% metal is lost, only 90% of the original volume is available for the new shapes. So, (Volume of new solid × number) = (100 - loss%)/100 × Volume of original solid. Always maintain consistent units and use the appropriate π value as specified in the question.

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Yes — this page has 44+ original Class 10 Mathematics Chapter 12 (“Surface Areas and Volumes”) important questions with answers (Multiple Choice (MCQ), Assertion–Reason, Short Answer, Short Answer, Long Answer, Case Study). Practise them free, or generate a full CBSE board-pattern sample paper (80 marks) and export it to PDF or Word — in English & Hindi, for 2026-27.

CBSE Class 10 Mathematics Chapter 12, Surface Areas and Volumes, elevates students' understanding from simple solids to composite shapes. This chapter focuses on three main areas: finding the surface area and volume of combinations of solids (like a toy made of a cone and hemisphere, or a tent combining a cylinder and a cone), converting one solid into another (recasting, with or without material loss), and calculating the curved surface area and volume of a frustum of a cone. The real challenge lies in visualising which surfaces are exposed or hidden when solids are combined, and applying formulas accurately. For example, a tent’s canvas excludes the base, while a toy’s total surface area might neglect the contacting faces. Exam questions often present practical problems: determining the number of small objects cast from a melted sphere, finding area of material needed for a structure, or computing volume of earth excavated to build an embankment. A deep grasp of formulas for cylinder, cone, sphere, and hemisphere, along with the frustum, is essential. Teachers and students should practice a variety of problems, paying special attention to unit consistency and the handling of π as either 22/7 or 3.14. This chapter tests calculation, spatial reasoning, and problem decomposition, making it a cornerstone for board exams.

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Mathematics — Surface Areas and Volumes

Class 10Time: 3 hrsMax Marks: 80

SECTION A

1.
A frustum of a cone has radii of its circular ends as 20 cm and 12 cm and slant height 17 cm. Its volume is:
(a) 3920π cm³(b) 3920 cm³(c) 1960π cm³(d) 1960 cm³
1
2.
The curved surface area of a frustum of a cone with end radii r₁, r₂ and slant height l is:
(a) π(r₁ + r₂)l(b) π(r₁ - r₂)l(c) 2π(r₁ + r₂)l(d) π(r₁² + r₂²)l
1
3.
A cylinder of radius R and height H is filled with water. A solid toy is made by attaching a cone to a hemisphere of same radius r (r < R). The total height of the toy is 2r. When the toy is completely submerged in the cylinder, the water level rises by:
(a) r³/R²(b) r³/(3R²)(c) 2r³/R²(d) 4r³/(3R²)
1

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Marks distribution & blueprint

In a CBSE exam, this chapter typically contributes questions across the following types. The last column shows how many original questions of each type we have ready in our bank for this chapter:

Question type	Marks each	In our bank
Multiple Choice (MCQ)	1 mark	13
Assertion–Reason	1 mark	6
Short Answer	2 marks	8
Short Answer	3 marks	6
Long Answer	5 marks	5
Case Study	4 marks	6

44 original, exam-style questions in our bank for this chapter — with answers.

Important & sample questions (with answers)

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Q1. A frustum of a cone has radii of its circular ends as 20 cm and 12 cm and slant height 17 cm. Its volume is:
1 mark
Multiple Choice (MCQ)
(A) 3920π cm³(B) 3920 cm³(C) 1960π cm³(D) 1960 cm³
▸ Answer▾ Answer
3920π cm³
Q2. The curved surface area of a frustum of a cone with end radii r₁, r₂ and slant height l is:
1 mark
Multiple Choice (MCQ)
(A) π(r₁ + r₂)l(B) π(r₁ - r₂)l(C) 2π(r₁ + r₂)l(D) π(r₁² + r₂²)l
▸ Answer▾ Answer
π(r₁ + r₂)l
Q3. A cylinder of radius R and height H is filled with water. A solid toy is made by attaching a cone to a hemisphere of same radius r (r < R). The total height of the toy is 2r. When the toy is completely submerged in the cylinder, the water level rises by:
1 mark
Multiple Choice (MCQ)
(A) r³/R²(B) r³/(3R²)(C) 2r³/R²(D) 4r³/(3R²)
▸ Answer▾ Answer
r³/R²
Q4. A cylindrical bucket of radius 10 cm and height 30 cm is filled with sand. The sand is emptied on the ground to form a conical heap of height 15 cm. The radius of the heap is:
1 mark
Multiple Choice (MCQ)
(A) 10√6 cm(B) 20√3 cm(C) 30 cm(D) √600 cm
▸ Answer▾ Answer
10√6 cm
Q5. Assertion (A): A toy is in the shape of a right circular cone mounted on a hemisphere. The radius of the hemisphere is 3.5 cm and the total height of the toy is 15.5 cm. The total surface area of the toy is 214.5 cm². Reason (R): The total surface area of the combined solid is the sum of the total surface areas of the individual solids.
1 mark
Assertion–Reason
(A) Both A and R are true and R is the correct explanation of A.(B) Both A and R are true but R is not the correct explanation of A.(C) A is true but R is false.(D) A is false but R is true.
▸ Answer▾ Answer
A is true but R is false.
Q6. If the ratio of the volume to the surface area of a sphere is 1 : 3, find the radius of the sphere.
2 marks
Short Answer
▸ Answer▾ Answer
The radius of the sphere is 1 unit.
Q7. A well of diameter 4 m is dug 21 m deep. The earth taken out of it is spread evenly all around it to form an embankment of width 2 m. Find the height of the embankment.
2 marks
Short Answer
▸ Answer▾ Answer
The height of the embankment is 7 m.
Q8. A solid is made by attaching a hemisphere to the larger base of a frustum of a cone. The hemisphere has radius 9 cm and exactly covers the larger base of the frustum. The smaller base radius of the frustum is 5 cm and the frustum height is 8 cm. Calculate the total surface area of the solid. (Use π = 3.14 and √5 = 2.236)
3 marks
Short Answer
▸ Answer▾ Answer
(187π + 56π√5) cm² or approximately 980.36 cm²
Q9. A bucket is in the form of the frustum of a cone. The radii of its circular ends are 28 cm and 21 cm, and its height is 24 cm. Find the capacity of the bucket in litres. (Use π = 22/7)
3 marks
Short Answer
▸ Answer▾ Answer
45.584 litres
Q10. A solid toy is in the shape of a right circular cone placed on a right circular cylinder. The common diameter is 6 cm. The height of the cylinder is 6 cm and the total height of the toy is 10 cm. Find the total surface area of the toy. (Use π = 3.14, and note that the base of the toy is flat and exposed.)
5 marks
Long Answer
▸ Answer▾ Answer
188.4 cm²
Q11. A solid iron rectangular block of dimensions 4.4 m, 2.6 m and 1 m is melted and cast into a hollow cylindrical pipe of internal radius 30 cm and thickness 5 cm. Find the length of the pipe in metres. (Use π = 22/7)
5 marks
Long Answer
▸ Answer▾ Answer
112 m
Q12. A camping tent is designed as a cylinder of diameter 6 m and height 4 m, surmounted by a cone of the same diameter such that the highest point of the tent is 7 m from the ground. The tent is made of canvas and sits on the ground, with no base canvas.
4 marks
Case Study
1. (i) Find the slant height of the conical part.1 mark
2. (ii) Calculate the area of canvas required to make the tent (excluding the base). (Take π = 3.14)2 marks
3. (iii) What is the volume of air inside the tent?1 mark
▸ Answer▾ Answer
Slant height = 3√2 m (≈4.24 m), Canvas area = 115.30 m², Volume = 141.30 m³

Frequently asked questions

What are the typical problem types in CBSE Class 10 Surface Areas and Volumes?

Exam questions usually fall into three categories: (1) finding surface area/volume of combined solids like toys, tents, or decorative items; (2) conversion of solids (melting and recasting), often with material loss; and (3) frustum of a cone, focusing on curved surface area, total surface area, and volume. Real-life contexts like wells, embankments, and canvas tents are common. Students must carefully note which surfaces are exposed and which are in contact to avoid adding extra areas.

How can students avoid mistakes when calculating surface areas of combined solids?