Exercise 12.1 answers: Surface Areas and Volumes

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1. Each cube has volume 64 cm³, so its edge is the cube root of 64, which is 4 cm.
2. Joining the two cubes end to end gives a cuboid of length 4+4 = 8 cm, breadth 4 cm and height 4 cm.
3. When the cubes join, the two faces that touch (each 4×4 = 16 cm²) go inside and stop counting as outer surface.
4. Surface area of the cuboid = 2×(length×breadth + breadth×height + height×length) = 2×(8×4 + 4×4 + 4×8) = 2×80 = 160 cm².

Answer160 cm²

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1. The hemisphere's diameter is 14 cm, so its radius is 7 cm — and this is also the cylinder's radius, since they join smoothly.
2. The hemisphere itself takes up a height of 7 cm, so the cylinder's height is 13 − 7 = 6 cm.
3. Since the vessel is hollow and open at the top, the inner surface = curved surface of the cylinder + curved surface of the hemisphere.
4. Inner surface area = 2πrh + 2πr² = 2πr(h+r) = 2×(22/7)×7×(6+7) = 2×22×13 = 572 cm².

Answer572 cm²

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1. The cone and hemisphere share radius 3.5 cm.
2. The hemisphere takes up 3.5 cm of the total height, so the cone's own height is 15.5 − 3.5 = 12 cm.
3. Slant height of the cone: l = √(r²+h²) = √(3.5² + 12²) = √156.25 = 12.5 cm.
4. Total surface area (the flat circle where they join is hidden) = curved surface of cone + curved surface of hemisphere = πrl + 2πr² = πr(l+2r) = (22/7)×3.5×(12.5+7) = 11×19.5 = 214.5 cm².

Answer214.5 cm²

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1. The hemisphere sits on the cube's flat face, so its circular base cannot be wider than the cube's edge.
2. The greatest diameter it can have is therefore equal to the cube's edge: 7 cm, giving radius 3.5 cm.
3. The hemisphere hides a circle of area πr² on the cube's top face, but adds curved surface 2πr² in its place — a net gain of πr².
4. Surface area of the solid = 6×(side)² + πr² = 6×49 + (22/7)×3.5² = 294 + 38.5 = 332.5 cm².

AnswerGreatest diameter = 7 cm; surface area of the solid = 332.5 cm²

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1. The cube has edge l, so its own surface area is 6l².
2. The hemispherical hollow has diameter l, so radius l/2.
3. Scooping it out removes the flat circle of area π(l/2)² from the top face, but the bowl-shaped hollow left behind has curved surface 2π(l/2)² — exactly double.
4. Surface area of the remaining solid = 6l² − π(l/2)² + 2π(l/2)² = 6l² + π(l/2)² = 6l² + πl²/4.

Answer6l² + πl²/4

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1. Diameter 5 mm means radius 2.5 mm, shared by the cylinder and both hemispherical ends.
2. The two hemispherical ends together take up 2×2.5 = 5 mm of the total length, so the cylindrical part is 14 − 5 = 9 mm long.
3. Both flat circles where the hemispheres meet the cylinder are hidden, so only curved surfaces are outside: the cylinder's curved wall plus two hemisphere domes (which together make one full sphere's curved surface).
4. Surface area = 2πrh + 4πr² = 2×(22/7)×2.5×9 + 4×(22/7)×2.5² = 141.43 + 78.57 = 220 mm² (using 2πr(h+2r) = 2×(22/7)×2.5×14 = 220 mm²).

Answer220 mm²

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1. Diameter 4 m means radius 2 m, shared by the cylinder and the cone on top.
2. The tent's canvas is only the curved parts — no floor and no hidden join circle: curved surface of cylinder + curved surface of cone.
3. Curved surface of cylinder = 2πrh = 2×(22/7)×2×2.1 = 26.4 m². Curved surface of cone = πrl = (22/7)×2×2.8 = 17.6 m².
4. Total canvas = 26.4 + 17.6 = 44 m². Cost = 44×₹500 = ₹22,000.

AnswerCanvas used = 44 m²; cost = ₹22,000

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1. Diameter 1.4 cm gives radius 0.7 cm, shared by the cylinder and the cone drilled out of it, and both have height 2.4 cm.
2. Slant height of the cone: l = √(r²+h²) = √(0.49+5.76) = √6.25 = 2.5 cm.
3. Since the cone's mouth exactly matches the cylinder's top, the whole top circle disappears — replaced by the cone's inward curved surface. The cylinder's curved wall and bottom circle stay as they were.
4. Total surface area = curved surface of cylinder + bottom circle + curved surface of cone = 2πrh + πr² + πrl = πr(2h+r+l) = (22/7)×0.7×(4.8+0.7+2.5) = 2.2×8 = 17.6 cm², which rounds to 18 cm².

Answer18 cm² (to the nearest cm²)

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1. The scooped hemispheres share the cylinder's own radius, 3.5 cm, since they are cut from its flat ends.
2. Each end's flat circle disappears completely (the hemisphere is exactly as wide as the cylinder) and is replaced by the hemisphere's curved bowl.
3. So the total surface = curved surface of the cylinder + two hemisphere curved surfaces (which together equal one full sphere's curved surface).
4. Total surface area = 2πrh + 4πr² = 2×(22/7)×3.5×10 + 4×(22/7)×3.5² = 220 + 154 = 374 cm².

Answer374 cm²

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