Surface area and volume of prisms, cylinders, right cones, spheres and right pyramids (and which height each one uses), composite solids built from them, and what happens to area and volume when every dimension is multiplied by a factor k.
Grade 10 adds the right cone, the sphere and the right pyramid to the prisms and cylinders you already know. It then asks you to work with composite solids — two of these joined together, or one cut out of another — and it asks one new question: what happens when you scale a solid up?
1The formulae
Cylinder: V = πr2h · SA = 2πr2 + 2πrh
Cone: V = ⅓πr2h · SA = πr2 + πrℓ
Sphere: V = 4/3πr3 · SA = 4πr2
Right pyramid: V = ⅓ × base area × h · SA = base area + area of the triangular faces
h is always the perpendicular height, measured straight up from the base; ℓ is the slant height, measured up the sloping side.
The cone and pyramid both carry a ⅓; the cylinder and prism do not. That single detail is the most common lost mark in this topic.
- 1V = πr²h = 3,14 × 25 × 12.Square the radius first.
- 2V = 942 cm³.
- 1V = ⅓πr²h = ⅓ × 3,14 × 9 × 4.
- 2V = ⅓ × 113,04 = 37,68 cm³.Exactly a third of the matching cylinder.
- 3Slant height: ℓ2 = 42 + 32 = 25, so ℓ = 5 cm.The height, the radius and the slant height form a right-angled triangle inside the cone.
- 4SA = πr2 + πrℓ = 3,14 × 9 + 3,14 × 3 × 5 = 28,26 + 47,10 = 75,36 cm².The flat base plus the curved surface — and the curved surface uses ℓ, never h.
- 1SA = 4πr² = 4 × 3,14 × 36.
- 2SA = 452,16 cm².
2Right pyramids: two different heights (Grade 10)
Grade 11, read this first. Pyramids are Grade 10 measurement. Your Term 4 revision list is: “Revise the volume and surface areas of right-prisms and cylinders”, the effect of multiplying a dimension by k, and “volume and surface areas of spheres, right prisms, right cones and combination of those objects” — the pyramid is not on it. Work through this section if you want the revision; it will not be set in a Grade 11 paper.
In a right pyramid the apex sits directly above the centre of the base. In Grade 10 the base is a square or an equilateral triangle. A pyramid has two heights, and every pyramid question turns on using the right one.
Surface area uses the slant height: SA = base area + area of the triangular faces
Square base of side b: SA = b2 + 4 × 1/2bℓ
From one height to the other: ℓ2 = h2 + (1/2b)2
- 1Slant height: ℓ2 = 42 + 32 = 25, so ℓ = 5 cm.Half the base edge is 3 cm. The height, the half-base and the slant height form a right-angled triangle.
- 2Base = 6 × 6 = 36 cm2. Each face = 1/2 × 6 × 5 = 15 cm2.Each face is a triangle with base 6 cm and height ℓ = 5 cm.
- 3SA = 36 + 4 × 15 = 96 cm2.One square base and four identical triangles.
- 4V = ⅓ × 36 × 4 = 48 cm3.Volume uses the perpendicular height, 4 cm — not the slant height.
Using the perpendicular height for the faces: 4 × 1/2 × 6 × 4 = 48 cm2 is too small, because the faces lean. The height of each face is the slant height ℓ = 5 cm, which gives 60 cm2. The rule of thumb: volume → h, surface area → ℓ.
Equilateral-triangle base? Find the height of the base triangle with Pythagoras first. For sides of 6 cm it is √(62 − 32) = √27 ≈ 5,20 cm, so the base area is 1/2 × 6 × √27 ≈ 15,59 cm2. Then add the three sloping faces, each 1/2 × 6 × ℓ.
3Composite solids
A composite solid is built from the solids above — a cylinder with a dome on top, a cone on a cylinder, a block with a hollow cut into it. There is no new formula, only three decisions:
- Split it into solids you know, and write down each part’s own measurements. A dome sitting exactly on a cylinder has the cylinder’s radius.
- Volume: ADD the parts that are there; SUBTRACT any part that has been cut out or hollowed.
- Surface area: count only the surfaces on the OUTSIDE of the finished solid. Where two parts are joined, that face is sealed inside — leave it out of both parts. Where a hollow is cut, its inside walls become new surface.
- 1V = πr2h + 2/3πr3 = π(3)2(10) + 2/3π(3)3A hemisphere is half a sphere: 1/2 × 4/3πr3 = 2/3πr3. Both parts are there, so ADD them.
- 2V = 90π + 18π = 108π ≈ 339,29 m3Keep π to the last line and round once, at the end.
- 3The surface is the flat base, the cylinder’s curved wall and the dome’s curved surface.The circle where the dome meets the cylinder (dashed) is sealed inside the solid. It is NOT surface — not as the top of the cylinder and not as the flat face of the hemisphere.
- 4SA = πr2 + 2πrh + 2πr2 = 9π + 60π + 18π = 87π ≈ 273,32 m2The curved surface of a hemisphere is half of 4πr2, which is 2πr2.
Adding the two solids’ own surface areas counts the joined circle twice — once as the top of the cylinder and once as the flat face of the hemisphere. Here that adds 2 × 9π ≈ 56,55 m2 of surface that does not exist.
- 1V = π(6)2(10) − 1/3π(6)2(8) = 360π − 96π = 264π ≈ 829,38 cm3The cone has been taken away, so SUBTRACT it.
- 2Slant height of the cone: ℓ2 = 82 + 62 = 100, so ℓ = 10 cm.The cone’s curved wall is now an inside surface of the solid, and its area needs ℓ.
- 3The surface is the base, the outer curved wall and the curved wall of the hollow.The top circle is gone — it is the open mouth of the hollow.
- 4SA = π(6)2 + 2π(6)(10) + π(6)(10) = 36π + 120π + 60π = 216π ≈ 678,58 cm2The curved surface of a cone is πrℓ.
4The effect of scaling
If every dimension of a solid is multiplied by k:
- 1k = 2, and volume scales by k³.Volume involves three dimensions.
- 22³ = 8, so the volume becomes 8 times as big.Its surface area becomes 2² = 4 times as big.
Doubling the dimensions does not double the volume — it multiplies it by 8. Area goes up by k² and volume by k³, never by k.
Practice exercises
Work each one out, then click to reveal the answer.
- 1A cube has a volume of 216 cm3. Determine the length of one edge, and hence the total surface area of the cube.
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edge = ∛216 = 6 cm (2)
A cube has 6 square faces, so SA = 6 × 62 (1)
= 216 cm2 (1) - 2The diagram below shows a cylinder. Calculate its volume. (π ≈ 3,14)
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3,14 × 25 × 12 = 942 cm³ - 3The diagram below shows a cone. Calculate its volume. (π ≈ 3,14)
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⅓ × 3,14 × 9 × 4 = 37,68 cm³ - 4Calculate the volume of the sphere in the diagram below. (π ≈ 3,14)
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4/3 × 3,14 × 27 = 113,04 cm³ - 5Calculate the surface area of the sphere in the diagram below. (π ≈ 3,14)
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4 × 3,14 × 36 = 452,16 cm² - 6The diagram below shows a pyramid. Calculate its volume.
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⅓ × 30 × 9 = 90 cm³ - 7The diagram below shows a right pyramid with a square base of side 18 cm and a perpendicular height of 12 cm. Calculate the slant height ℓ and the total surface area of the pyramid.
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Half the base edge = 9 cm; ℓ2 = 122 + 92 = 225, so ℓ = 15 cm (2)
Base = 182 = 324 cm2 (1); four faces = 4 × 1/2 × 18 × 15 = 540 cm2 (1)
TSA = 324 + 540 = 864 cm2 (1) - 8The diagram below shows a right pyramid whose base is an equilateral triangle with sides of 8 cm. The slant height of each sloping face is 10 cm. Calculate the total surface area of the pyramid.
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Height of the base triangle = √(82 − 42) = √48 ≈ 6,93 cm (1)
Base area = 1/2 × 8 × √48 ≈ 27,71 cm2 (1)
Three faces = 3 × 1/2 × 8 × 10 = 120 cm2 (1)
TSA ≈ 147,71 cm2 (1) - 9A solid's dimensions are all doubled. What happens to its volume?
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It becomes 8 times bigger (2³). - 10A solid's dimensions are all tripled. What happens to its surface area?
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It becomes 9 times bigger (3²). - 11The diagram below shows a triangular prism. Calculate its volume.
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½ × 6 × 4 = 12 cm²; 12 × 10 = 120 cm³ - 12Calculate the volume of the cylinder in the diagram below. (π ≈ 3,14)
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3,14 × 16 × 8 = 401,92 cm³ - 13The solid in the diagram below is a cylinder of radius 4 cm and height 10 cm with a cone of height 3 cm on top. Calculate the volume and the total surface area of the solid, correct to TWO decimal places.
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V = π(4)2(10) + 1/3π(4)2(3) = 160π + 16π = 176π ≈ 552,92 cm3 (2)
Slant height of the cone: ℓ = √(32 + 42) = 5 cm (1)
TSA = base + curved wall + curved cone = π(4)2 + 2π(4)(10) + π(4)(5) (1)
= 16π + 80π + 20π = 116π ≈ 364,42 cm2 (1)
The circle where the cone meets the cylinder is sealed inside and is not counted (1) - 14A cube with edges of 3 cm is glued to the centre of the top of a cube with edges of 6 cm, as in the diagram below. Calculate the total surface area of the solid.
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Large cube: 6 × 62 = 216 cm2; small cube: 6 × 32 = 54 cm2 (1)
The joined 3 cm by 3 cm square is hidden on BOTH cubes: subtract 2 × 9 = 18 cm2 (1)
TSA = 216 + 54 − 18 = 252 cm2 (1) - 15Which two solids have a ⅓ in their volume formula?
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The cone and the pyramid
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