Question 1
Two cubes of volume 64 cm³ are joined face to face. Find the surface area of the resulting cuboid.
Steps only
Cube side = ∛64 = 4 cm
Cuboid dimensions = 8 × 4 × 4
Surface area = 2(8×4 + 4×4 + 8×4) = 160
Answer
160 cm².
CLASS 10 · MATHEMATICS · CHAPTER 12
Surface Areas and Volumes
Break combined solids into familiar shapes, count only exposed surfaces and add or subtract volumes as the situation requires.
17 worked answersAll 2 exercise sets
What you will learn
- Identify exposed surfaces after solids are joined or hollowed.
- Add and subtract volumes of cones, cylinders and spheres.
- Keep dimensions and units consistent.
17 worked answers · short steps first, explanations when you need them
Exercise 12.1
Textbook pages 166–167
Question 2
A hollow hemisphere of diameter 14 cm is topped by a hollow cylinder. The total vessel height is 13 cm. Find its inner surface area.
Steps only
r = 7 cm; cylinder height = 13 − 7 = 6 cm
Area = 2πr² + 2πrh = 2π×7(7+6) = 572
Answer
572 cm².
Question 3
A cone of radius 3.5 cm is mounted on a hemisphere of the same radius. Total height is 15.5 cm. Find the toy's total surface area.
Steps only
Cone height = 15.5 − 3.5 = 12 cm; slant height = √(12² + 3.5²) = 12.5 cm
Area = πrl + 2πr² = π×3.5(12.5+7) = 214.5
Answer
214.5 cm².
Question 4
A cube of side 7 cm is surmounted by the largest possible hemisphere. Find its diameter and the solid's surface area.
Steps only
Largest diameter = cube side = 7 cm; r = 3.5 cm
Area = 6×7² − πr² + 2πr² = 294 + 38.5 = 332.5
Answer
Diameter 7 cm; surface area 332.5 cm².
Question 5
A hemispherical depression of diameter l is cut from one face of a cube of edge l. Find the remaining surface area.
Steps only
r = l/2
Area = 6l² − πr² + 2πr² = 6l² + πl²/4
Answer
l²(24 + π)/4 square units.
Question 6
A capsule is a cylinder with two hemispherical ends. Its total length is 14 mm and diameter 5 mm. Find its surface area.
Steps only
r = 2.5 mm; cylinder height = 14 − 5 = 9 mm
Area = 2πrh + 4πr² = 45π + 25π = 220
Answer
220 mm².
Question 7
A tent has a cylinder of height 2.1 m and diameter 4 m, topped by a cone of slant height 2.8 m. Find canvas area and cost at ₹500 per m².
Steps only
r = 2 m; canvas = 2πrh + πrl
= π×2(2×2.1 + 2.8) = 44 m²
Cost = 44 × 500 = ₹22,000
Answer
44 m²; ₹22,000.
Question 8
A conical cavity is hollowed from a cylinder of height 2.4 cm and diameter 1.4 cm, with the same height and diameter. Find the remaining total surface area to the nearest cm².
Steps only
r = 0.7 cm; cone slant height = √(2.4² + 0.7²) = 2.5 cm
Area = 2πrh + πr² + πrl = πr(2h + r + l) = 17.6
Answer
18 cm² to the nearest cm².
Question 9
A cylinder of height 10 cm and radius 3.5 cm has a hemisphere scooped from each end. Find its total surface area.
Steps only
Area = cylinder curved area + two hemispherical curved areas
= 2πrh + 4πr² = 70π + 49π = 374
Answer
374 cm².
Exercise 12.2
Textbook pages 169–170
Question 1
A cone stands on a hemisphere. Both radii are 1 cm and the cone height is 1 cm. Find the volume in terms of π.
Steps only
Volume = 2/3πr³ + 1/3πr²h
= 2π/3 + π/3 = π
Answer
π cm³.
Question 2
A 12 cm model consists of a cylinder and two end cones, each 2 cm high. Its diameter is 3 cm. Find the enclosed air volume.
Steps only
r = 1.5 cm; cylinder height = 12 − 2 − 2 = 8 cm
Volume = πr²×8 + 2(1/3πr²×2) = 21π = 66
Answer
66 cm³.
Question 3
Each of 45 gulab jamuns is a 5 cm capsule of diameter 2.8 cm. Syrup fills about 30% of its volume. Find the total syrup volume.
Steps only
r = 1.4 cm; cylinder length = 5 − 2.8 = 2.2 cm
One volume = πr²h + 4/3πr³ ≈ 25.05 cm³
Syrup = 45 × 0.30 × 25.05 ≈ 338.2
Answer
Approximately 338 cm³.
Question 4
A 15 cm × 10 cm × 3.5 cm cuboid has four conical holes of radius 0.5 cm and depth 1.4 cm. Find the wood volume.
Steps only
Cuboid volume = 15×10×3.5 = 525
Four holes = 4×(1/3×22/7×0.5²×1.4) = 22/15
Wood = 525 − 22/15 = 523.53
Answer
Approximately 523.53 cm³.
Question 5
A cone of height 8 cm and top radius 5 cm is full of water. Spherical lead shots of radius 0.5 cm displace one fourth of the water. Find the number of shots.
Steps only
Displaced volume = 1/4×1/3π×5²×8 = 50π/3
One shot = 4/3π×0.5³ = π/6
Number = (50π/3)/(π/6) = 100
Answer
100 lead shots.
Question 6
An iron pole has a cylinder of height 220 cm and diameter 24 cm, topped by a cylinder of height 60 cm and radius 8 cm. Find its mass if iron is 8 g/cm³, using π = 3.14.
Steps only
Volume = 3.14(12²×220 + 8²×60) = 111532.8 cm³
Mass = 111532.8×8 = 892262.4 g
Answer
892.2624 kg.
Question 7
A cone of height 120 cm and radius 60 cm stands on a radius-60 cm hemisphere inside a full cylinder of radius 60 cm and height 180 cm. Find the water left.
Steps only
Cylinder = π×60²×180 = 648000π
Solid = 1/3π×60²×120 + 2/3π×60³ = 288000π
Water left = 360000π
Answer
360000π cm³ = 7920000/7 cm³.
Question 8
A spherical vessel has an 8 cm cylindrical neck of diameter 2 cm and a spherical part of diameter 8.5 cm. A child reports 345 cm³. Check using π = 3.14.
Steps only
Sphere radius = 4.25 cm; neck radius = 1 cm
Volume = 4/3×3.14×4.25³ + 3.14×1²×8 ≈ 346.51 cm³
Answer
No. The calculated capacity is about 346.51 cm³.