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NCERT Class 10 Mathematics Exercise 12.1 Solutions: Surface Areas and Volumes
AI-written, checked on 23 Oct 2026
Solutions written by Shishya, checked — not the board's official answers
Step-by-step working for every question of Exercise 12.1 in NCERT's Class 10 Mathematics book, chapter “Surface Areas and Volumes”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.
How these were checked: the solutions were drafted with AI and then read and corrected on 23 Oct 2026. Every numerical final answer is recomputed from the question's own numbers by a calculator check that runs before each release; proofs and explanations were read, not machine-checked. The NCERT textbook is the authority — if something here disagrees with it, follow the book.
Tap “why?” under a step to see the textbook rule it uses — from Shishya's fixed list of NCERT Class 9-10 rules, not written by AI for this page. A step without one has no rule from the list yet.
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Ex 12.1 Q1
Two cubes of volume 64 cm³ each are joined end to end. Find the surface area of the cuboid formed.
- Let the side of each cube be a cm.
- Volume of one cube is a³ = 64 cm³.
why?
Volumes: cylinder πr²h, cone ⅓πr²h, sphere ⁴⁄₃πr³, hemisphere ⅔πr³, cuboid l × b × h. The volume of a solid made of parts is the sum of the parts' volumes; melting or recasting keeps the volume the same.
- Solving for a, we get a = 4 cm.
- When two cubes are joined end to end, the dimensions of the cuboid are length 8 cm, breadth 4 cm, and height 4 cm.
- Surface area of the cuboid is 2(lb + bh + hl).
why?
Curved surface areas: cylinder 2πrh, cone πrl (slant height l = √(r² + h²)), sphere 4πr², hemisphere 2πr². When solids are joined, only the faces still on the outside are counted.
- Substituting the values, SA = 2(8 × 4 + 4 × 4 + 4 × 8).
why?
Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
- SA = 2(32 + 16 + 32).
- SA = 2(80) = 160 cm².
Answer: 160 cm²
Ex 12.1 Q2
A vessel is a hollow hemisphere of diameter 14 cm topped by a hollow cylinder; total height 13 cm. Find the inner surface area (π = 22/7).
- Given diameter of the hemisphere is 14 cm, so its radius r is 7 cm.
- The total height of the vessel is 13 cm.
- The height of the cylindrical part is the total height minus the radius of the hemisphere, which is 13 − 7 = 6 cm.
- The inner surface area consists of the curved surface area of the hemisphere and the curved surface area of the cylinder.
why?
Curved surface areas: cylinder 2πrh, cone πrl (slant height l = √(r² + h²)), sphere 4πr², hemisphere 2πr². When solids are joined, only the faces still on the outside are counted.
- Curved surface area of the hemisphere is 2πr² = 2 × (22/7) × 7² = 308 cm².
why?
Curved surface areas: cylinder 2πrh, cone πrl (slant height l = √(r² + h²)), sphere 4πr², hemisphere 2πr². When solids are joined, only the faces still on the outside are counted.
- Curved surface area of the cylinder is 2πrh = 2 × (22/7) × 7 × 6 = 264 cm².
why?
Curved surface areas: cylinder 2πrh, cone πrl (slant height l = √(r² + h²)), sphere 4πr², hemisphere 2πr². When solids are joined, only the faces still on the outside are counted.
- Total inner surface area is the sum of the two areas: 308 + 264 = 572 cm².
why?
Curved surface areas: cylinder 2πrh, cone πrl (slant height l = √(r² + h²)), sphere 4πr², hemisphere 2πr². When solids are joined, only the faces still on the outside are counted.
Answer: 572 cm²
Ex 12.1 Q3
A toy is a cone of radius 3.5 cm on a hemisphere of the same radius; total height 15.5 cm. Find its total surface area (π = 22/7).
- Given radius of cone and hemisphere r = 3.5 cm and total height of toy = 15.5 cm.
- Height of the cone h = Total height − Radius of hemisphere = 15.5 − 3.5 = 12 cm.
- Slant height of the cone l = √(r² + h²) = √(3.5² + 12²) = √(12.25 + 144) = √156.25 = 12.5 cm.
why?
Curved surface areas: cylinder 2πrh, cone πrl (slant height l = √(r² + h²)), sphere 4πr², hemisphere 2πr². When solids are joined, only the faces still on the outside are counted.
- Total surface area of the toy = Curved surface area of cone + Curved surface area of hemisphere.
why?
Curved surface areas: cylinder 2πrh, cone πrl (slant height l = √(r² + h²)), sphere 4πr², hemisphere 2πr². When solids are joined, only the faces still on the outside are counted.
- Total surface area = πrl + 2πr² = πr(l + 2r).
why?
Curved surface areas: cylinder 2πrh, cone πrl (slant height l = √(r² + h²)), sphere 4πr², hemisphere 2πr². When solids are joined, only the faces still on the outside are counted.
- Substituting values, Total surface area = (22/7) × 3.5 × (12.5 + 2 × 3.5).
why?
Curved surface areas: cylinder 2πrh, cone πrl (slant height l = √(r² + h²)), sphere 4πr², hemisphere 2πr². When solids are joined, only the faces still on the outside are counted.
- Total surface area = 11 × (12.5 + 7) = 11 × 19.5 = 214.5 cm².
why?
Curved surface areas: cylinder 2πrh, cone πrl (slant height l = √(r² + h²)), sphere 4πr², hemisphere 2πr². When solids are joined, only the faces still on the outside are counted.
Answer: 214.5 cm²
Ex 12.1 Q4
A hemisphere sits on a cube of side 7 cm. Find the greatest diameter the hemisphere can have and the surface area of the solid (π = 22/7).
- The hemisphere sits on a face of the cube, so its diameter can be at most the cube's side: greatest diameter = 7 cm, radius r = 3.5 cm.
- Surface area of the solid = surface area of the cube − area of the circle covered by the hemisphere + curved surface of the hemisphere.
why?
Curved surface areas: cylinder 2πrh, cone πrl (slant height l = √(r² + h²)), sphere 4πr², hemisphere 2πr². When solids are joined, only the faces still on the outside are counted.
- = 6 × 7² − πr² + 2πr² = 294 + πr².
why?
Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.
- πr² = (22/7) × 3.5 × 3.5 = 38.5 cm².
why?
Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.
- Surface area = 294 + 38.5 = 332.5 cm².
why?
Curved surface areas: cylinder 2πrh, cone πrl (slant height l = √(r² + h²)), sphere 4πr², hemisphere 2πr². When solids are joined, only the faces still on the outside are counted.
Answer: Greatest diameter 7 cm; surface area 332.5 cm²
Ex 12.1 Q5
A hemisphere of diameter l (the cube's edge) is scooped out of one face of a cube. Find the surface area of the remaining solid.
- The cube has an edge length of l.
- The hemisphere has a diameter of l, so its radius r is l/2.
- The surface area of the cube is 6l².
why?
Curved surface areas: cylinder 2πrh, cone πrl (slant height l = √(r² + h²)), sphere 4πr², hemisphere 2πr². When solids are joined, only the faces still on the outside are counted.
- A hemisphere is scooped out from one face, removing a circular area of πr².
why?
Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.
- The area of the removed circular face is π(l/2)², which is πl²/4.
- The curved surface area of the hemisphere is added to the solid, which is 2πr².
why?
Curved surface areas: cylinder 2πrh, cone πrl (slant height l = √(r² + h²)), sphere 4πr², hemisphere 2πr². When solids are joined, only the faces still on the outside are counted.
- The area of the curved surface is 2π(l/2)², which is πl²/2.
why?
Curved surface areas: cylinder 2πrh, cone πrl (slant height l = √(r² + h²)), sphere 4πr², hemisphere 2πr². When solids are joined, only the faces still on the outside are counted.
- The total surface area is the cube's area minus the removed circle plus the curved surface.
why?
Curved surface areas: cylinder 2πrh, cone πrl (slant height l = √(r² + h²)), sphere 4πr², hemisphere 2πr². When solids are joined, only the faces still on the outside are counted.
- Total Surface Area = 6l² − πl²/4 + πl²/2.
why?
Curved surface areas: cylinder 2πrh, cone πrl (slant height l = √(r² + h²)), sphere 4πr², hemisphere 2πr². When solids are joined, only the faces still on the outside are counted.
- Simplifying the expression gives 6l² + πl²/4.
why?
Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
- This can be written as l²(24 + π)/4.
Answer: l²(24 + π)/4
Ex 12.1 Q6
A capsule is a cylinder with a hemisphere on each end; total length 14 mm, diameter 5 mm. Find its surface area (π = 22/7).
Uses a figure printed in the NCERT book — open the chapter above to see it.
- The capsule consists of a cylinder with a hemisphere attached at each end.
- Given total length of capsule = 14 mm and diameter = 5 mm.
- Radius of capsule r = diameter ÷ 2 = 5 ÷ 2 = 2.5 mm.
- Length of the cylindrical part h = Total length − 2 × radius = 14 − 2 × 2.5 = 14 − 5 = 9 mm.
- Surface area of capsule = Curved surface area of cylinder + Surface area of two hemispheres.
why?
Curved surface areas: cylinder 2πrh, cone πrl (slant height l = √(r² + h²)), sphere 4πr², hemisphere 2πr². When solids are joined, only the faces still on the outside are counted.
- Surface area of two hemispheres = 4πr² = 4 × (22/7) × (2.5)² = 4 × (22/7) × 6.25 = 79.14 mm².
why?
Curved surface areas: cylinder 2πrh, cone πrl (slant height l = √(r² + h²)), sphere 4πr², hemisphere 2πr². When solids are joined, only the faces still on the outside are counted.
- Curved surface area of cylinder = 2πrh = 2 × (22/7) × 2.5 × 9 = 141.36 mm².
why?
Curved surface areas: cylinder 2πrh, cone πrl (slant height l = √(r² + h²)), sphere 4πr², hemisphere 2πr². When solids are joined, only the faces still on the outside are counted.
- Total surface area = 79.14 + 141.36 = 220.5 mm².
why?
Curved surface areas: cylinder 2πrh, cone πrl (slant height l = √(r² + h²)), sphere 4πr², hemisphere 2πr². When solids are joined, only the faces still on the outside are counted.
- Using π = 22/7 exactly: 4 × (22/7) × (5/2)² + 2 × (22/7) × (5/2) × 9 = 4 × (22/7) × (25/4) + 2 × (22/7) × (45/2) = 22 × 25/7 + 22 × 45/7 = 550/7 + 990/7 = 1540/7 = 220 mm².
Answer: 220 mm²
Ex 12.1 Q7
A tent is a cylinder (height 2.1 m, diameter 4 m) topped by a cone of slant height 2.8 m. Find the canvas area (no base) and its cost at ₹500 per m² (π = 22/7).
- The diameter of the cylinder is 4 m, so its radius r is 4 ÷ 2 = 2 m.
- The height of the cylindrical part h is 2.1 m.
- The slant height of the conical part l is 2.8 m.
why?
Curved surface areas: cylinder 2πrh, cone πrl (slant height l = √(r² + h²)), sphere 4πr², hemisphere 2πr². When solids are joined, only the faces still on the outside are counted.
- The canvas area is the sum of the curved surface area of the cylinder and the curved surface area of the cone.
why?
Curved surface areas: cylinder 2πrh, cone πrl (slant height l = √(r² + h²)), sphere 4πr², hemisphere 2πr². When solids are joined, only the faces still on the outside are counted.
- Curved surface area of the cylinder = 2πrh = 2 × (22/7) × 2 × 2.1 = 26.4 m².
why?
Curved surface areas: cylinder 2πrh, cone πrl (slant height l = √(r² + h²)), sphere 4πr², hemisphere 2πr². When solids are joined, only the faces still on the outside are counted.
- Curved surface area of the cone = πrl = (22/7) × 2 × 2.8 = 17.6 m².
why?
Curved surface areas: cylinder 2πrh, cone πrl (slant height l = √(r² + h²)), sphere 4πr², hemisphere 2πr². When solids are joined, only the faces still on the outside are counted.
- Total canvas area = 26.4 + 17.6 = 44 m².
- Cost of canvas = Area × Rate per m² = 44 × 500 = ₹22000.
Answer: Canvas area = 44 m²; Cost = ₹22000
Ex 12.1 Q8
A cone of the same height (2.4 cm) and diameter (1.4 cm) is hollowed out of a solid cylinder. Find the total surface area of the remaining solid to the nearest cm² (π = 22/7).
- The radius of the cylinder and cone is r = 1.4 ÷ 2 = 0.7 cm.
- The height of the cylinder and cone is h = 2.4 cm.
- The slant height of the cone is l = √(r² + h²) = √(0.7² + 2.4²) = √(0.49 + 5.76) = √6.25 = 2.5 cm.
why?
Curved surface areas: cylinder 2πrh, cone πrl (slant height l = √(r² + h²)), sphere 4πr², hemisphere 2πr². When solids are joined, only the faces still on the outside are counted.
- The total surface area of the remaining solid is the curved surface area of the cylinder plus the top and bottom circular faces minus the base of the cone plus the curved surface area of the cone.
why?
Curved surface areas: cylinder 2πrh, cone πrl (slant height l = √(r² + h²)), sphere 4πr², hemisphere 2πr². When solids are joined, only the faces still on the outside are counted.
- This simplifies to 2πrh + 2πr² − πr² + πrl = 2πrh + πr² + πrl.
why?
Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.
- Substituting the values, TSA = πr(2h + r + l) = (22/7) × 0.7 × (2 × 2.4 + 0.7 + 2.5).
why?
Curved surface areas: cylinder 2πrh, cone πrl (slant height l = √(r² + h²)), sphere 4πr², hemisphere 2πr². When solids are joined, only the faces still on the outside are counted.
- Calculating the term in brackets: 4.8 + 0.7 + 2.5 = 8.0.
- Multiplying the terms: (22/7) × 0.7 × 8.0 = 22 × 0.1 × 8.0 = 17.6 cm².
- Rounding 17.6 cm² to the nearest cm² gives 18 cm².
Answer: 17.6 cm², 18 cm²
Ex 12.1 Q9
A hemisphere is scooped out of each end of a solid cylinder of height 10 cm and radius 3.5 cm. Find the total surface area of the article (π = 22/7).
Uses a figure printed in the NCERT book — open the chapter above to see it.
- The article consists of a cylindrical part with two hemispherical ends removed.
- The total surface area is the sum of the curved surface area of the cylinder and the curved surface areas of the two hemispheres.
why?
Curved surface areas: cylinder 2πrh, cone πrl (slant height l = √(r² + h²)), sphere 4πr², hemisphere 2πr². When solids are joined, only the faces still on the outside are counted.
- Radius r = 3.5 cm and height of the cylinder h = 10 cm.
- Curved surface area of the cylinder = 2πrh = 2 × (22/7) × 3.5 × 10 = 220 cm².
why?
Curved surface areas: cylinder 2πrh, cone πrl (slant height l = √(r² + h²)), sphere 4πr², hemisphere 2πr². When solids are joined, only the faces still on the outside are counted.
- Curved surface area of one hemisphere = 2πr² = 2 × (22/7) × 3.5 × 3.5 = 77 cm².
why?
Curved surface areas: cylinder 2πrh, cone πrl (slant height l = √(r² + h²)), sphere 4πr², hemisphere 2πr². When solids are joined, only the faces still on the outside are counted.
- Curved surface area of two hemispheres = 2 × 77 = 154 cm².
why?
Curved surface areas: cylinder 2πrh, cone πrl (slant height l = √(r² + h²)), sphere 4πr², hemisphere 2πr². When solids are joined, only the faces still on the outside are counted.
- Total surface area = Curved surface area of cylinder + Curved surface area of two hemispheres.
why?
Curved surface areas: cylinder 2πrh, cone πrl (slant height l = √(r² + h²)), sphere 4πr², hemisphere 2πr². When solids are joined, only the faces still on the outside are counted.
- Total surface area = 220 + 154 = 374 cm².
why?
Curved surface areas: cylinder 2πrh, cone πrl (slant height l = √(r² + h²)), sphere 4πr², hemisphere 2πr². When solids are joined, only the faces still on the outside are counted.
Answer: 374 cm²
Other exercises in this chapter
- Exercise 12.2 — 8 questions
← Surface Areas and Volumes: notes, practice and all exercises
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