Class 10 Maths Chapter 12 Surface Areas and Volumes Extra Questions (HOTS)

These HOTS-level questions go beyond Exercise 12.1/12.2 — general/abstract reasoning, reverse-engineering, and fresh numbers not used in the Solutions post, designed for students aiming for full marks on combination-of-solids problems. These Class 10 Mathematics Chapter 12 important questions are handy for last-minute exam practice.

Last Updated: September 10, 2026

Q1 (General derivation). A solid consists of a right circular cylinder of radius r and height h, with a hemisphere of the same radius attached to one end and a cone of the same radius and height h₁ attached to the other end. Derive a general formula for the total surface area of the solid in terms of r, h and h₁.
Solution: Slant height of cone, l = √(r²+h₁²). The two curved ends are the hemisphere’s curved surface and the cone’s curved surface; the cylinder contributes only its curved surface (both circular ends are covered by the attached solids).
TSA = (curved surface of cylinder) + (curved surface of hemisphere) + (curved surface of cone) = 2πrh + 2πr² + πrl = πr(2h + 2r + √(r²+h₁²)).

Q2 (Reverse-engineering). A solid toy is in the shape of a hemisphere surmounted by a cone of the same radius. If the total surface area of the toy is 235.5 cm² and the radius is 3.5 cm, find the height of the cone.
Solution: TSA = πr(l+2r) ⟹ 235.5 = (22/7)(3.5)(l+7) ⟹ 235.5 = 11(l+7) ⟹ l+7 = 21.409 ⟹ l ≈ 14.409 cm.
h = √(l²−r²) = √(207.6 − 12.25) = √195.35 ≈ 13.98 cm ≈ 14 cm.

Q3 (Assertion-Reason). Assertion (A): When a solid cone is carved out of a solid hemisphere of the same base radius, the surface area of the remaining solid equals its original curved surface area plus the cone’s curved surface area.
Reason (R): Removing the cone exposes a new conical cavity whose curved surface becomes part of the outer boundary, while the flat circular base remains unaffected.
Solution: Both A and R are true, and R correctly explains A — the flat base stays as-is, only the top curved surface changes from a flat circle to the cone cavity’s slant surface. Answer: (a) Both A and R are true, and R is the correct explanation of A.

Q4 (Fresh-numbers word problem). A cylindrical tub of radius 6 cm and height 15 cm is full of water. A solid in the shape of a cone mounted on a hemisphere (both of radius 3 cm and the cone’s height 6 cm) is dropped into the tub. Find the volume of water that overflows.
Solution: Volume of solid = (1/3)π(3²)(6) + (2/3)π(3³) = 18π + 18π = 36π cm³ = 36×22/7 ≈ 113.14 cm³.
Since the tub was full, the water that overflows equals the volume of the submerged solid = ≈113.14 cm³.

Q5 (Ratio/general proof). A sphere of radius R is melted and recast into a right circular cone of the same radius R. Prove that the height of the cone is 4R, and find the ratio of the cone’s slant height to R.
Solution: Volume conserved: (4/3)πR³ = (1/3)πR²h ⟹ h = 4R (proved).
Slant height l = √(R²+16R²) = √17·R. Ratio l : R = √17 : 1 (≈ 4.123 : 1).

Q6 (Multi-solid word problem, fresh numbers). A juice glass is a frustum-free combination: a cylinder of radius 3.5 cm and height 10 cm topped by a hemisphere of the same radius used as a lid. If the cylinder is filled with juice to 80% of its volume, find the volume of juice.
Solution: Cylinder volume = πr²h = (22/7)(12.25)(10) = 385 cm³. 80% of this = 0.8×385 = 308 cm³. (The hemisphere lid is not filled — it is a cover, not a liquid-holding region, testing careful reading of “which part holds liquid.”)

More on This Chapter

• Solutions
• Revision Notes
• Formulas Handbook (Class 10 Maths)
• Book Catalog Page

Written by Satish

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