April 10, 2026
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Chapter 12: Surface Areas and Volumes
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Exercise 12.1 (Page 244)
13 marks
2 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.
Answer:
Volume of cube = 64 cm³ → side = 4 cm.
Two cubes joined → length = 4+4 = 8 cm, breadth = 4 cm, height = 4 cm.
Surface area = 2(lb + bh + hl) = 2(8×4 + 4×4 + 4×8) = 2(32+16+32) = 2×80 = 160 cm².
Surface area = 160 cm².
Volume of cube = 64 cm³ → side = 4 cm.
Two cubes joined → length = 4+4 = 8 cm, breadth = 4 cm, height = 4 cm.
Surface area = 2(lb + bh + hl) = 2(8×4 + 4×4 + 4×8) = 2(32+16+32) = 2×80 = 160 cm².
Surface area = 160 cm².
23 marks
A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Answer:
Diameter = 14 cm → radius r = 7 cm. Height of cylinder = total height - radius of hemisphere = 13 - 7 = 6 cm.
Inner surface area = CSA of hemisphere + CSA of cylinder = 2πr² + 2πrh = 2πr(r + h) = 2×(22/7)×7×(7+6) = 44 × 13 = 572 cm².
Inner surface area = 572 cm².
Diameter = 14 cm → radius r = 7 cm. Height of cylinder = total height - radius of hemisphere = 13 - 7 = 6 cm.
Inner surface area = CSA of hemisphere + CSA of cylinder = 2πr² + 2πrh = 2πr(r + h) = 2×(22/7)×7×(7+6) = 44 × 13 = 572 cm².
Inner surface area = 572 cm².
33 marks
A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Answer:
Radius r = 3.5 cm = 7/2 cm. Height of cone = total height - radius = 15.5 - 3.5 = 12 cm.
Slant height l = √(r² + h²) = √((7/2)² + 12²) = √(49/4 + 144) = √((49+576)/4) = √(625/4) = 25/2 = 12.5 cm.
Surface area = CSA of cone + CSA of hemisphere = πrl + 2πr² = πr(l + 2r) = (22/7)×(7/2)×(12.5 + 7) = 11 × 19.5 = 214.5 cm².
Total surface area = 214.5 cm².
Radius r = 3.5 cm = 7/2 cm. Height of cone = total height - radius = 15.5 - 3.5 = 12 cm.
Slant height l = √(r² + h²) = √((7/2)² + 12²) = √(49/4 + 144) = √((49+576)/4) = √(625/4) = 25/2 = 12.5 cm.
Surface area = CSA of cone + CSA of hemisphere = πrl + 2πr² = πr(l + 2r) = (22/7)×(7/2)×(12.5 + 7) = 11 × 19.5 = 214.5 cm².
Total surface area = 214.5 cm².
43 marks
A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.
Answer:
Side of cube = 7 cm. Hemisphere can have maximum diameter = 7 cm → radius = 3.5 cm.
Surface area = TSA of cube + CSA of hemisphere - area of circular base (which is covered) = 6×7² + 2πr² - πr² = 294 + πr² = 294 + (22/7)×(3.5)² = 294 + (22/7)×12.25 = 294 + 38.5 = 332.5 cm².
Surface area = 332.5 cm².
Side of cube = 7 cm. Hemisphere can have maximum diameter = 7 cm → radius = 3.5 cm.
Surface area = TSA of cube + CSA of hemisphere - area of circular base (which is covered) = 6×7² + 2πr² - πr² = 294 + πr² = 294 + (22/7)×(3.5)² = 294 + (22/7)×12.25 = 294 + 38.5 = 332.5 cm².
Surface area = 332.5 cm².
53 marks
A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
Answer:
Let edge of cube = l, radius of hemisphere = l/2.
Surface area = TSA of cube - area of circular face + CSA of hemisphere = 6l² - π(l/2)² + 2π(l/2)² = 6l² - πl²/4 + 2πl²/4 = 6l² + πl²/4 = l²(6 + π/4).
Surface area = l²(6 + π/4).
Let edge of cube = l, radius of hemisphere = l/2.
Surface area = TSA of cube - area of circular face + CSA of hemisphere = 6l² - π(l/2)² + 2π(l/2)² = 6l² - πl²/4 + 2πl²/4 = 6l² + πl²/4 = l²(6 + π/4).
Surface area = l²(6 + π/4).
63 marks
A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see Fig. 12.2). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
Answer:
Diameter = 5 mm → radius r = 2.5 mm. Total length = 14 mm. Cylinder length = 14 - 2×2.5 = 9 mm.
Surface area = CSA of cylinder + CSA of two hemispheres = 2πrh + 2×2πr² = 2πr(h + 2r) = 2×(22/7)×2.5×(9 + 5) = (110/7)×14 = 220 mm².
Surface area = 220 mm².
Diameter = 5 mm → radius r = 2.5 mm. Total length = 14 mm. Cylinder length = 14 - 2×2.5 = 9 mm.
Surface area = CSA of cylinder + CSA of two hemispheres = 2πrh + 2×2πr² = 2πr(h + 2r) = 2×(22/7)×2.5×(9 + 5) = (110/7)×14 = 220 mm².
Surface area = 220 mm².
73 marks
A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas at the rate of ₹ 500 per m².
Answer:
Radius = 2 m. Cylinder height = 2.1 m, cone slant height = 2.8 m.
Canvas area = CSA of cylinder + CSA of cone = 2πrh + πrl = 2π×2×2.1 + π×2×2.8 = 8.4π + 5.6π = 14π = 14×3.14 = 43.96 m².
Cost = 43.96 × 500 = ₹ 21980.
Area = 44 m² (approx.), Cost = ₹ 21980.
Radius = 2 m. Cylinder height = 2.1 m, cone slant height = 2.8 m.
Canvas area = CSA of cylinder + CSA of cone = 2πrh + πrl = 2π×2×2.1 + π×2×2.8 = 8.4π + 5.6π = 14π = 14×3.14 = 43.96 m².
Cost = 43.96 × 500 = ₹ 21980.
Area = 44 m² (approx.), Cost = ₹ 21980.
83 marks
From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid.
Answer:
Radius r = 0.7 cm, height h = 2.4 cm. Slant height of cone l = √(r²+h²) = √(0.49+5.76)=√6.25=2.5 cm.
TSA of remaining = CSA of cylinder + CSA of cone + area of base (circular rim) = 2πrh + πrl + πr² = πr(2h + l + r).
= (22/7)×0.7×(4.8+2.5+0.7) = 2.2×8 = 17.6 cm².
Total surface area = 17.6 cm².
Radius r = 0.7 cm, height h = 2.4 cm. Slant height of cone l = √(r²+h²) = √(0.49+5.76)=√6.25=2.5 cm.
TSA of remaining = CSA of cylinder + CSA of cone + area of base (circular rim) = 2πrh + πrl + πr² = πr(2h + l + r).
= (22/7)×0.7×(4.8+2.5+0.7) = 2.2×8 = 17.6 cm².
Total surface area = 17.6 cm².
93 marks
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. 12.7. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.
Answer:
Radius r = 3.5 cm, cylinder height = 10 cm. Hemispheres of same radius are scooped out from both ends.
TSA = CSA of cylinder + CSA of two hemispheres = 2πrh + 2×2πr² = 2πr(h + 2r) = 2×(22/7)×3.5×(10+7) = 22 × 17 = 374 cm².
Total surface area = 374 cm².
Radius r = 3.5 cm, cylinder height = 10 cm. Hemispheres of same radius are scooped out from both ends.
TSA = CSA of cylinder + CSA of two hemispheres = 2πrh + 2×2πr² = 2πr(h + 2r) = 2×(22/7)×3.5×(10+7) = 22 × 17 = 374 cm².
Total surface area = 374 cm².
Exercise 12.2 (Page 247) – Volume of Combined Solids
13 marks
A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.
Answer:
Radius r = 1 cm, cone height = r = 1 cm.
Volume = Volume of cone + Volume of hemisphere = (1/3)πr²h + (2/3)πr³ = (1/3)π×1×1 + (2/3)π×1 = (π/3) + (2π/3) = π cm³.
Volume = π cm³.
Radius r = 1 cm, cone height = r = 1 cm.
Volume = Volume of cone + Volume of hemisphere = (1/3)πr²h + (2/3)πr³ = (1/3)π×1×1 + (2/3)π×1 = (π/3) + (2π/3) = π cm³.
Volume = π cm³.
23 marks
Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model.
Answer:
Diameter = 3 cm → radius = 1.5 cm. Total length = 12 cm. Each cone height = 2 cm, so cylinder height = 12 - 4 = 8 cm.
Volume = Volume of cylinder + 2×Volume of cone = πr²h_cyl + 2×(1/3)πr²h_cone = π×1.5²×8 + (2/3)π×1.5²×2 = π×2.25×8 + (2/3)π×2.25×2 = 18π + 3π = 21π cm³.
Volume = 21π cm³ ≈ 66 cm³.
Diameter = 3 cm → radius = 1.5 cm. Total length = 12 cm. Each cone height = 2 cm, so cylinder height = 12 - 4 = 8 cm.
Volume = Volume of cylinder + 2×Volume of cone = πr²h_cyl + 2×(1/3)πr²h_cone = π×1.5²×8 + (2/3)π×1.5²×2 = π×2.25×8 + (2/3)π×2.25×2 = 18π + 3π = 21π cm³.
Volume = 21π cm³ ≈ 66 cm³.
33 marks
A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see Fig. 12.15).
Answer:
Diameter = 2.8 cm → radius = 1.4 cm. Total length = 5 cm. Cylinder length = 5 - 2×1.4 = 2.2 cm.
Volume of one jamun = Volume of cylinder + Volume of sphere (two hemispheres) = πr²h + (4/3)πr³ = π×1.4²×2.2 + (4/3)π×1.4³ = π(1.96×2.2 + (4/3)×2.744) = π(4.312 + 3.6587) = 7.9707π cm³ ≈ 25.05 cm³.
Syrup in one = 30% of 25.05 = 7.515 cm³. For 45 jamuns = 45×7.515 = 338.175 cm³ ≈ 338 cm³.
Approx syrup = 338 cm³.
Diameter = 2.8 cm → radius = 1.4 cm. Total length = 5 cm. Cylinder length = 5 - 2×1.4 = 2.2 cm.
Volume of one jamun = Volume of cylinder + Volume of sphere (two hemispheres) = πr²h + (4/3)πr³ = π×1.4²×2.2 + (4/3)π×1.4³ = π(1.96×2.2 + (4/3)×2.744) = π(4.312 + 3.6587) = 7.9707π cm³ ≈ 25.05 cm³.
Syrup in one = 30% of 25.05 = 7.515 cm³. For 45 jamuns = 45×7.515 = 338.175 cm³ ≈ 338 cm³.
Approx syrup = 338 cm³.
43 marks
A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm × 10 cm × 3.5 cm. The radius of each depression is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand.
Answer:
Volume of cuboid = 15×10×3.5 = 525 cm³.
Volume of one cone = (1/3)πr²h = (1/3)×(22/7)×0.5²×1.4 = (1/3)×(22/7)×0.25×1.4 = (1/3)×(22/7)×0.35 = (1/3)×(22×0.05) = (1/3)×1.1 = 0.3667 cm³.
Four cones = 4×0.3667 = 1.4668 cm³.
Wood volume = 525 - 1.4668 = 523.5332 cm³ ≈ 523.53 cm³.
Volume of wood = 523.53 cm³.
Volume of cuboid = 15×10×3.5 = 525 cm³.
Volume of one cone = (1/3)πr²h = (1/3)×(22/7)×0.5²×1.4 = (1/3)×(22/7)×0.25×1.4 = (1/3)×(22/7)×0.35 = (1/3)×(22×0.05) = (1/3)×1.1 = 0.3667 cm³.
Four cones = 4×0.3667 = 1.4668 cm³.
Wood volume = 525 - 1.4668 = 523.5332 cm³ ≈ 523.53 cm³.
Volume of wood = 523.53 cm³.
53 marks
A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm, are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped.
Answer:
Volume of cone = (1/3)πr²h = (1/3)π×5²×8 = (200π)/3 cm³.
Water overflow = 1/4 of cone volume = (1/4)×(200π/3) = (50π)/3 cm³.
Volume of one lead shot (sphere) = (4/3)πr³ = (4/3)π×(0.5)³ = (4/3)π×0.125 = (0.5π)/3 cm³.
Number of shots = (Volume overflow) / (Volume per shot) = [(50π)/3] / [(0.5π)/3] = 50/0.5 = 100.
Number of lead shots = 100.
Volume of cone = (1/3)πr²h = (1/3)π×5²×8 = (200π)/3 cm³.
Water overflow = 1/4 of cone volume = (1/4)×(200π/3) = (50π)/3 cm³.
Volume of one lead shot (sphere) = (4/3)πr³ = (4/3)π×(0.5)³ = (4/3)π×0.125 = (0.5π)/3 cm³.
Number of shots = (Volume overflow) / (Volume per shot) = [(50π)/3] / [(0.5π)/3] = 50/0.5 = 100.
Number of lead shots = 100.
63 marks
A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm³ of iron has approximately 8 g mass. (Use π = 3.14)
Answer:
For larger cylinder: radius = 12 cm, height = 220 cm → volume = π×12²×220 = 3.14×144×220 = 3.14×31680 = 99475.2 cm³.
For smaller cylinder: radius = 8 cm, height = 60 cm → volume = 3.14×64×60 = 3.14×3840 = 12057.6 cm³.
Total volume = 99475.2 + 12057.6 = 111532.8 cm³.
Mass = 111532.8 × 8 = 892262.4 g = 892.26 kg.
Mass ≈ 892.26 kg.
For larger cylinder: radius = 12 cm, height = 220 cm → volume = π×12²×220 = 3.14×144×220 = 3.14×31680 = 99475.2 cm³.
For smaller cylinder: radius = 8 cm, height = 60 cm → volume = 3.14×64×60 = 3.14×3840 = 12057.6 cm³.
Total volume = 99475.2 + 12057.6 = 111532.8 cm³.
Mass = 111532.8 × 8 = 892262.4 g = 892.26 kg.
Mass ≈ 892.26 kg.
73 marks
A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.
Answer:
Volume of cylinder = πr²h = π×60²×180 = π×3600×180 = 648000π cm³.
Volume of solid = cone + hemisphere = (1/3)π×60²×120 + (2/3)π×60³ = (1/3)π×3600×120 + (2/3)π×216000 = 144000π + 144000π = 288000π cm³.
Water left = 648000π - 288000π = 360000π cm³ = 360000×3.14 = 1130400 cm³ = 1.1304 m³.
Volume left = 1.1304 m³ (approx).
Volume of cylinder = πr²h = π×60²×180 = π×3600×180 = 648000π cm³.
Volume of solid = cone + hemisphere = (1/3)π×60²×120 + (2/3)π×60³ = (1/3)π×3600×120 + (2/3)π×216000 = 144000π + 144000π = 288000π cm³.
Water left = 648000π - 288000π = 360000π cm³ = 360000×3.14 = 1130400 cm³ = 1.1304 m³.
Volume left = 1.1304 m³ (approx).
83 marks
A spherical glass vessel has a cylindrical neck 8 cm long and 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm³. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.
Answer:
Sphere radius = 4.25 cm, volume = (4/3)πr³ = (4/3)×3.14×76.7656 ≈ 321.39 cm³.
Cylinder radius = 1 cm, height = 8 cm, volume = πr²h = 3.14×1×8 = 25.12 cm³.
Total volume = 321.39 + 25.12 = 346.51 cm³ ≈ 347 cm³. Child's 345 cm³ is close but slightly less; due to rounding, she is approximately correct.
Approx correct (346.5 cm³).
Sphere radius = 4.25 cm, volume = (4/3)πr³ = (4/3)×3.14×76.7656 ≈ 321.39 cm³.
Cylinder radius = 1 cm, height = 8 cm, volume = πr²h = 3.14×1×8 = 25.12 cm³.
Total volume = 321.39 + 25.12 = 346.51 cm³ ≈ 347 cm³. Child's 345 cm³ is close but slightly less; due to rounding, she is approximately correct.
Approx correct (346.5 cm³).
Key Concepts Summary
Formula 1Surface Areas
List the surface area formulas for common solids.
Answer: Cube: 6a²; Cuboid: 2(lb+bh+hl); Cylinder: 2πrh + 2πr²; Cone: πrl + πr²; Sphere: 4πr²; Hemisphere: 3πr² (total) or 2πr² (curved).
Formula 2Volumes
List the volume formulas for common solids.
Answer: Cube: a³; Cuboid: lbh; Cylinder: πr²h; Cone: (1/3)πr²h; Sphere: (4/3)πr³; Hemisphere: (2/3)πr³.
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