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Class 10th Mathematics Chapter 12 Surface Areas and Volumes – Important Questions

Important questions with answers for Class 10th Mathematics Chapter 12 Surface Areas and Volumes, based on the NCERT CBSE syllabus.

Topics covered in Surface Areas and Volumes

Frequently asked questions

Where can I find important questions for Class 10th Mathematics Chapter 12 Surface Areas and Volumes?

This page lists important questions with answers for Surface Areas and Volumes. Topics covered: Combination of solids, Frustum of cone, Conversion of solids.

Are Surface Areas and Volumes important questions free?

Yes, Notebook gives free NCERT CBSE important questions for Class 10 Mathematics.

How to prepare Surface Areas and Volumes for Class 10 Mathematics exams?

Read the Surface Areas and Volumes notes first, revise the key topics (Combination of solids, Frustum of cone, Conversion of solids), then practise the important questions and previous year questions (PYQ) for this chapter.

Where can I find Class 10th Mathematics Chapter 12 Surface Areas and Volumes important questions for quick revision?

You are on the right page: this important questions page for Surface Areas and Volumes covers Class 10 Mathematics for the CBSE exam 2026-27. Use the tabs on this chapter for notes, important questions and PYQ.

Is Class 10th Mathematics Chapter 12 Surface Areas and Volumes important for the CBSE exam 2026-27?

Surface Areas and Volumes is part of the Class 10 Mathematics NCERT CBSE syllabus. Notebook gives notes, important questions and PYQs for it, free.

Also see: Class 10th Mathematics Chapter 12 Surface Areas and Volumes Notes · Class 10th Mathematics Chapter 12 Surface Areas and Volumes PYQ

Surface Areas and Volumes

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3 Marks
A cone of radius 6 cm and height 8 cm is melted and recast into a sphere. Find the radius of the sphere.
Ans: Volume of cone = (1/3)π(36)(8) = 96π cm³. Volume of sphere = (4/3)πr³ = 96π → r³ = 72 → r = ∛72 ≈ 4.16 cm.
5 Marks
A wooden toy is in the shape of a cone mounted on a cylinder. Find its total surface area if the cylinder has radius 3 cm and height 5 cm, and the cone has slant height 5 cm.
Ans: TSA = curved SA of cone + curved SA of cylinder + base area of cylinder = πrl + 2πrh + πr² = π×3×5 + 2π×3×5 + π×9 = 15π + 30π + 9π = 54π ≈ 169.6 cm².
3 Marks
Define frustum of a cone and write its volume formula.
Ans: A frustum is the portion of a cone between two parallel planes cutting it. Volume = (πh/3)(R² + r² + Rr), where R and r are radii of the two circular ends and h is the height.
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