CBSE • Class 10 • Mathematics • Chapter 12 (Surface Areas and Volumes) • Exercise 12.1

Exercise 12.1: Surface Areas and Volumes — NCERT Solutions

Surface area of a combination of solids — cone on a hemisphere, cuboid with a cylinder, and similar.

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What this exercise covers

Combined surface areaCone + hemisphereCylinder + coneReal-life shapes

Step-by-step solutions — Exercise 12.1

6 solved questions • Each solution includes a Socratic hint, full working and a common-mistake callout • Last reviewed 2026-09-03

Exercise 12.1 Q1 • 2 marks

Two cubes each of volume 64 cm364\ \text{cm}^3 are joined end to end. Find the surface area of the resulting cuboid.
Hint (Socratic — try this first)▾
If you place two identical cubes side by side, how many square faces get hidden inside the joint?
Step-by-step solution▾

Volume of each cube =64 cm3= 64\ \text{cm}^3, so edge a=643=4 cma = \sqrt[3]{64} = 4\ \text{cm}.

Joining two cubes end to end gives a cuboid with:

  • length l=4+4=8 cml = 4+4 = 8\ \text{cm}
  • breadth b=4 cmb = 4\ \text{cm}
  • height h=4 cmh = 4\ \text{cm}

Surface area of cuboid: 2(lb+bh+hl)=2(8×4+4×4+4×8)2(lb+bh+hl) = 2(8\times4 + 4\times4 + 4\times8) =2(32+16+32)=2×80=160 cm2= 2(32+16+32) = 2\times80 = 160\ \text{cm}^2

So the surface area is 160 cm2160\ \text{cm}^2.

Common mistake:
Adding the two cubes' individual surface areas (6a2×2=1926a^2 \times 2 = 192) without subtracting the two hidden faces that are joined together.
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Exercise 12.1 Q2 • 3 marks

A toy is in the form of a cone of radius 3.5 cm3.5\ \text{cm} mounted on a hemisphere of the same radius. The total height of the toy is 15.5 cm15.5\ \text{cm}. Find the total surface area of the toy. (Take π=227\pi = \tfrac{22}{7})
Hint (Socratic — try this first)▾
Which surfaces are actually visible — do you use the flat base of the hemisphere or its curved surface?
Step-by-step solution▾

Radius r=3.5 cmr = 3.5\ \text{cm}.

Height of hemisphere =r=3.5 cm= r = 3.5\ \text{cm}. Height of cone h=15.5−3.5=12 cmh = 15.5 - 3.5 = 12\ \text{cm}.

Slant height of cone: l=r2+h2=3.52+122=12.25+144=156.25=12.5 cml = \sqrt{r^2 + h^2} = \sqrt{3.5^2 + 12^2} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5\ \text{cm}

Total surface area = CSA of cone + CSA of hemisphere: =πrl+2πr2=πr(l+2r)= \pi r l + 2\pi r^2 = \pi r(l + 2r) =227×3.5×(12.5+7)= \frac{22}{7}\times 3.5 \times (12.5 + 7) =227×3.5×19.5=11×19.5=214.5 cm2= \frac{22}{7}\times 3.5 \times 19.5 = 11 \times 19.5 = 214.5\ \text{cm}^2

Total surface area =214.5 cm2= 214.5\ \text{cm}^2.

Common mistake:
Including the flat circular base of the hemisphere in the surface area — it is joined to the cone and not exposed.
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Exercise 12.1 Q3 • 3 marks

A cubical block of side 7 cm7\ \text{cm} is surmounted by a hemisphere of the largest possible diameter. Find the surface area of the solid. (Take π=227\pi = \tfrac{22}{7})
Hint (Socratic — try this first)▾
The largest hemisphere sits on the top face — how does adding its curved surface change the exposed top face of the cube?
Step-by-step solution▾

Side of cube a=7 cma = 7\ \text{cm}.

Largest hemisphere has diameter equal to the side =7 cm= 7\ \text{cm}, so radius r=3.5 cmr = 3.5\ \text{cm}.

Surface area of solid = Surface area of cube − circular base area of hemisphere + curved surface of hemisphere: =6a2−πr2+2πr2=6a2+πr2= 6a^2 - \pi r^2 + 2\pi r^2 = 6a^2 + \pi r^2

Compute: 6a2=6×49=294 cm26a^2 = 6\times 49 = 294\ \text{cm}^2 πr2=227×3.5×3.5=38.5 cm2\pi r^2 = \frac{22}{7}\times 3.5 \times 3.5 = 38.5\ \text{cm}^2

Total =294+38.5=332.5 cm2= 294 + 38.5 = 332.5\ \text{cm}^2.

Common mistake:
Forgetting to subtract the circular base πr2\pi r^2 where the hemisphere covers the top face, so the top area is counted twice.
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Exercise 12.1 Q4 • 3 marks

A vessel is in the shape of a hollow cylinder mounted on a hollow hemisphere of the same radius 7 cm7\ \text{cm}. The total height of the vessel is 13 cm13\ \text{cm}. Find the inner surface area of the vessel. (Take π=227\pi = \tfrac{22}{7})
Hint (Socratic — try this first)▾
Which two curved surfaces line the inside — and what is the cylinder's height once the hemisphere's part is removed?
Step-by-step solution▾

Radius r=7 cmr = 7\ \text{cm}.

Height of hemisphere =r=7 cm= r = 7\ \text{cm}. Height of cylinder h=13−7=6 cmh = 13 - 7 = 6\ \text{cm}.

Inner surface area = CSA of cylinder + CSA of hemisphere: =2πrh+2πr2=2πr(h+r)= 2\pi r h + 2\pi r^2 = 2\pi r (h + r) =2×227×7×(6+7)= 2\times \frac{22}{7}\times 7 \times (6 + 7) =2×22×13=572 cm2= 2\times 22 \times 13 = 572\ \text{cm}^2

Inner surface area =572 cm2= 572\ \text{cm}^2.

Common mistake:
Using the total height of 13 cm13\ \text{cm} for the cylinder instead of subtracting the hemisphere's radius first.
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Exercise 12.1 Q5 • 3 marks

A medicine capsule is in the shape of a cylinder with two hemispheres stuck to its ends. The length of the entire capsule is 14 mm14\ \text{mm} and the diameter of the capsule is 5 mm5\ \text{mm}. Find its surface area. (Take π=227\pi = \tfrac{22}{7})
Hint (Socratic — try this first)▾
Two hemispheres of equal radius together make up how much of the two ends of the cylinder's length?
Step-by-step solution▾

Diameter =5 mm= 5\ \text{mm}, so radius r=2.5 mmr = 2.5\ \text{mm}.

The two hemispheres cover the ends, occupying r+r=5 mmr + r = 5\ \text{mm} of the length. Cylinder length h=14−5=9 mmh = 14 - 5 = 9\ \text{mm}.

Surface area = CSA of cylinder + 2 × CSA of hemisphere: =2πrh+2(2πr2)=2πrh+4πr2=2πr(h+2r)= 2\pi r h + 2(2\pi r^2) = 2\pi r h + 4\pi r^2 = 2\pi r(h + 2r) =2×227×2.5×(9+5)= 2\times \frac{22}{7}\times 2.5 \times (9 + 5) =2×227×2.5×14=2×22×2.5×2=220 mm2= 2\times \frac{22}{7}\times 2.5 \times 14 = 2\times 22 \times 2.5 \times 2 = 220\ \text{mm}^2

Surface area =220 mm2= 220\ \text{mm}^2.

Common mistake:
Subtracting only one radius (2.5 mm) from the length instead of the combined 5 mm for the two hemispherical ends.
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Exercise 12.1 Q6 • 3 marks

A tent is in the shape of a cylinder surmounted by a conical top. The radius of the base is 4 m4\ \text{m}, the height of the cylindrical part is 2.1 m2.1\ \text{m} and the slant height of the cone is 2.8 m2.8\ \text{m}. Find the area of the canvas required for the tent. (Take π=227\pi = \tfrac{22}{7})
Hint (Socratic — try this first)▾
Canvas covers only the walls and the sloping roof — is the flat base or the top circle ever included?
Step-by-step solution▾

Radius r=4 mr = 4\ \text{m}, cylinder height h=2.1 mh = 2.1\ \text{m}, cone slant height l=2.8 ml = 2.8\ \text{m}.

Canvas area = CSA of cylinder + CSA of cone: =2πrh+πrl=πr(2h+l)= 2\pi r h + \pi r l = \pi r(2h + l) =227×4×(2×2.1+2.8)= \frac{22}{7}\times 4 \times (2\times 2.1 + 2.8) =227×4×(4.2+2.8)= \frac{22}{7}\times 4 \times (4.2 + 2.8) =227×4×7=88 m2= \frac{22}{7}\times 4 \times 7 = 88\ \text{m}^2

Area of canvas required =88 m2= 88\ \text{m}^2.

Common mistake:
Including the base circle of the cylinder in the canvas area, or using the cone's vertical height instead of its slant height.
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How to approach Exercise 12.1

  1. Re-read the chapter summary first. Open Surface Areas and Volumes and refresh the key concepts: Cube, Cuboid, Sphere, Hemisphere.
  2. Try each problem yourself before opening the solution. Ten focused minutes per problem usually beats reading two finished solutions.
  3. Use Guru AI for the questions you get stuck on. The Socratic AI tutor walks you through it with hints instead of dictating the answer.
  4. Mark the questions you got wrong and revisit them after 24 hours — the spacing is what locks the method into long-term memory.

All exercises in Surface Areas and Volumes

  1. Exercise 12.1 — Surface area of a combination of solids — cone on a hemisphere, cuboid with a cylinder, and similar.
  2. Exercise 12.2 — Volume of a combination of solids and conversion-of-shape problems.

Exercise 12.2 →

Volume of a combination of solids and conversion-of-shape problems.

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