Revision notes, worked problems, a question bank, assertion and reason questions and answers to the NCERT exercises. Tap a group to open it; opening one closes the others.
| Solid | Curved / lateral surface area | Total surface area | Volume |
|---|---|---|---|
| Cuboid l × b × h | 2h(l + b) | 2(lb + bh + hl) | lbh |
| Cube of edge a | 4a² | 6a² | a³ |
| Cylinder (r, h) | 2πrh | 2πr(r + h) | πr²h |
| Cone (r, h, l) | πrl | πr(l + r) | ⅓πr²h |
| Sphere (r) | 4πr² | 4πr² | ⁴⁄₃πr³ |
| Hemisphere (r) | 2πr² | 3πr² | ⅔πr³ |
The total surface area of a combined solid is the sum of the curved surface areas of the parts that remain visible, plus any flat faces that are still exposed. Faces that are glued together are not counted.
A bird-bath is a cylinder of height 1.45 m and radius 30 cm with a hemispherical depression at the top. Its total surface area = CSA of cylinder + CSA of hemisphere:
When a solid is scooped out of another, the flat face disappears but the curved surface of the cavity is added. Total surface area can increase even though volume decreases.
The volume of a solid made by joining two solids is the sum of their volumes. If one is scooped out of another, subtract.
A toy is a cone (height 2 cm) on a hemisphere, diameter 4 cm. With π = 3.14:
A cylinder that just encloses it has r = 2 cm and h = 4 cm: 3.14 × 4 × 4 = 50.24 cm³. The difference is 25.12 cm³.
Number of small objects that can be made (or dropped) = (volume available) ÷ (volume of one object). Example: ¼ of a cone of radius 5 cm and height 8 cm holds 100 lead shots of radius 0.5 cm.
| Solid | Curved / lateral surface area | Total surface area | Volume |
|---|---|---|---|
| Cuboid l × b × h | 2h(l + b) | 2(lb + bh + hl) | lbh |
| Cube of edge a | 4a² | 6a² | a³ |
| Cylinder (r, h) | 2πrh | 2πr(r + h) | πr²h |
| Cone (r, h, l) | πrl | πr(l + r) | ⅓πr²h |
| Sphere (r) | 4πr² | 4πr² | ⁴⁄₃πr³ |
| Hemisphere (r) | 2πr² | 3πr² | ⅔πr³ |
Try each problem first, then open the solution to check your working.
Three cubes of edge 5 cm are joined end to end. Find the surface area of the cuboid.
Given: 15 cm × 5 cm × 5 cm
Use: 2(lb + bh + hl)
2(75 + 25 + 75) = 350
Answer: 350 cm²
A solid is a cylinder (r = 7 cm, h = 10 cm) with a cone of the same radius and height 24 cm on top. Find its total surface area.
Given: l = √(49 + 576) = 25
Use: πrl + 2πrh + πr²
227 × 7 × (25 + 20 + 7) = 22 × 52 = 1144
Answer: 1144 cm²
A toy is a cone of radius 7 cm and height 24 cm on a hemisphere of the same radius. Find its surface area.
Given: l = 25
Use: πrl + 2πr²
227 × 7 × (25 + 14) = 22 × 39 = 858
Answer: 858 cm²
A hemisphere of radius 3.5 cm sits on a cube of edge 10 cm. Find the total surface area.
Use: 6a² + πr²
600 + 227 × 12.25 = 600 + 38.5
Answer: 638.5 cm²
A capsule is a cylinder 10 mm long with hemispheres of radius 3.5 mm at both ends. Find its surface area.
Use: 2πrh + 4πr²
2 × 227 × 3.5 × (10 + 7) = 22 × 17
Answer: 374 mm²
A tent is a cylinder (radius 7 m, height 3 m) with a conical top of slant height 10 m. Find the canvas needed (no base).
Use: 2πrh + πrl
227 × 7 × (6 + 10) = 22 × 16 = 352
Answer: 352 m²
From a solid cylinder of height 8 cm and radius 6 cm, a cone of the same height and radius is removed. Find the total surface area of the remaining solid (π = 3.14).
Given: l = √(36 + 64) = 10
Use: 2πrh + πr² + πrl
3.14 × 6 × (16 + 6 + 10) = 3.14 × 192 = 602.88
Answer: 602.88 cm²
A toy is a cone of radius 3.5 cm and height 6 cm on a hemisphere of the same radius. Find its volume.
Use: ⅓πr²h + ⅔πr³
⅓ × 227 × 12.25 × (6 + 7) = 22 × 12.25 × 1321 ≈ 166.83
Answer: ≈ 166.83 cm³
A solid is a cylinder (r = 3 cm, h = 10 cm) with a hemisphere at each end. Find its volume (π = 3.14).
Use: πr²h + ⁴⁄₃πr³
3.14 × 9 × 10 + 4/3 × 3.14 × 27 = 282.6 + 113.04
Answer: 395.64 cm³
A shed is a cuboid 7 m × 15 m × 8 m with a half cylinder of diameter 7 m and length 15 m on top. Find its volume.
cuboid = 840
half cylinder = ½ × 227 × 12.25 × 15 = 288.75
total = 1128.75
Answer: 1128.75 m³
A cone of radius 3 cm and height 4 cm is placed in a cylinder of radius 3 cm full of water, height 10 cm. How much water remains? (π = 3.14)
Use: cylinder − cone
cylinder = 3.14 × 9 × 10 = 282.6
cone = ⅓ × 3.14 × 9 × 4 = 37.68
282.6 − 37.68
Answer: 244.92 cm³
How many lead balls of radius 1 cm can be made from a sphere of radius 8 cm?
Use: ratio of volumes
(8/1)³ = 512
Answer: 512
A pipe is 21 cm long with inner and outer radii 3 cm and 4 cm. Find the volume of metal.
Use: π(R² − r²)h
227 × 7 × 21 = 462
Answer: 462 cm³
A cylindrical bucket of radius 14 cm and height 30 cm is full of water. Its water is poured into conical cups of radius 7 cm and height 6 cm. How many cups can be filled?
bucket = π × 196 × 30 = 5880π
cup = ⅓ × π × 49 × 6 = 98π
5880 ÷ 98
Answer: 60 cups
A cylinder just encloses a hemisphere of radius 6 cm (same base, height 6 cm). Find the volume of the space between them (π = 3.14).
Use: πr²h − ⅔πr³
3.14 × 36 × 6 − ⅔ × 3.14 × 216 = 678.24 − 452.16
Answer: 226.08 cm³
A solid iron cylinder has radius 7 cm and height 10 cm. If 1 cm³ of iron weighs 8 g, find its mass.
V = 227 × 49 × 10 = 1540 cm³
mass = 1540 × 8 = 12320 g
Answer: 12.32 kg
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Answers to every exercise question of the NCERT chapter. Try each one first, then tap Show answer.
Edge of each cube = ∛64 = 4 cm. The cuboid is 8 cm × 4 cm × 4 cm.
Surface area = 2(8 × 4 + 4 × 4 + 4 × 8) = 2(32 + 16 + 32) = 160 cm².
Answer: 160 cm²
r = 7 cm; height of the cylinder = 13 − 7 = 6 cm.
Inner surface = 2πr² + 2πrh = 2πr(r + h) = 2 × 227 × 7 × 13 = 572 cm².
Answer: 572 cm²
Height of cone = 15.5 − 3.5 = 12 cm. l = √(12² + 3.5²) = √156.25 = 12.5 cm.
TSA = πrl + 2πr² = πr(l + 2r) = 227 × 3.5 × (12.5 + 7) = 11 × 19.5 = 214.5 cm².
Answer: 214.5 cm²
The hemisphere must fit on a 7 cm face, so the greatest diameter is 7 cm (r = 3.5 cm).
Surface area = 6 × 49 − πr² + 2πr² = 294 + πr² = 294 + 227 × 12.25 = 294 + 38.5 = 332.5 cm².
Answer: 7 cm; 332.5 cm²
r = l/2. Surface area = 6l² − πr² + 2πr² = 6l² + πr² = 6l² + πl²/4.
Answer: ¼ l²(π + 24)
r = 2.5 mm; cylinder length = 14 − 5 = 9 mm.
Surface area = 2πrh + 4πr² = 2πr(h + 2r) = 2 × 227 × 2.5 × 14 = 220 mm².
Answer: 220 mm²
r = 2 m. Canvas = 2πrh + πrl = πr(2h + l) = 227 × 2 × (4.2 + 2.8) = 227 × 14 = 44 m².
Cost = 44 × 500 = ₹22000.
Answer: 44 m²; ₹22000
r = 0.7 cm, h = 2.4 cm, l = √(2.4² + 0.7²) = 2.5 cm.
TSA = 2πrh + πr² (bottom) + πrl (cavity) = πr(2h + r + l) = 227 × 0.7 × (4.8 + 0.7 + 2.5) = 2.2 × 8 = 17.6 cm².
Answer: 17.6 cm² ≈ 18 cm²
TSA = 2πrh + 2 × 2πr² = 2πr(h + 2r) = 2 × 227 × 3.5 × (10 + 7) = 22 × 17 = 374 cm².
Answer: 374 cm²
V = ⅔πr³ + ⅓πr²h = ⅔π + ⅓π = π cm³.
Answer: π cm³
r = 1.5 cm; cylinder length = 12 − 4 = 8 cm.
V = πr²(8) + 2 × ⅓πr²(2) = πr²(8 + 43) = 227 × 2.25 × 283 = 66 cm³.
Answer: 66 cm³
r = 1.4 cm; cylinder length = 5 − 2.8 = 2.2 cm.
One gulab jamun = πr²(2.2) + ⁴⁄₃πr³ = 227 × 1.96 × 2.2 + ⁴⁄₃ × 227 × 2.744 ≈ 13.55 + 11.50 = 25.05 cm³.
Syrup in 45 = 45 × 25.05 × 0.3 ≈ 338.2 cm³.
Answer: about 338 cm³
Cuboid = 15 × 10 × 3.5 = 525 cm³.
Each cone = ⅓ × 227 × 0.25 × 1.4 = 1130 ≈ 0.367 cm³; four cones ≈ 1.467 cm³.
Wood = 525 − 1.467 ≈ 523.53 cm³.
Answer: ≈ 523.53 cm³
Cone = ⅓ × 227 × 25 × 8 = 440021 cm³. Water out = ¼ of this = 110021 cm³.
One shot = ⁴⁄₃ × 227 × (0.5)³ = 1121 cm³.
Number = 110021 ÷ 1121 = 100.
Answer: 100
V = 3.14 × 12² × 220 + 3.14 × 8² × 60 = 99475.2 + 12057.6 = 111532.8 cm³.
Mass = 111532.8 × 8 = 892262.4 g ≈ 892.26 kg.
Answer: ≈ 892.26 kg
Cylinder = π × 3600 × 180 = 648000π cm³.
Solid = ⅓π × 3600 × 120 + ⅔π × 60³ = 144000π + 144000π = 288000π cm³.
Water left = 360000π = 227 × 360000 ≈ 1131428.57 cm³ ≈ 1.131 m³.
Answer: ≈ 1.131 m³
Sphere: ⁴⁄₃ × 3.14 × 4.25³ ≈ 321.39 cm³. Neck: 3.14 × 1² × 8 = 25.12 cm³.
Total ≈ 346.51 cm³.
Answer: She is not exactly correct; the volume is about 346.51 cm³