Geometry & Measures
Grade 6-8Volume of Spheres, Cones & Pyramids
Spheres, cones and pyramids each have their own volume formula, given on the exam formula sheet, but examiners expect you to select the correct one, apply it accurately, and combine shapes together in composite solid questions. This lesson covers the volume and surface area formulae for spheres, cones and pyramids, combining two solids such as a cone topped with a hemisphere, and working backwards from a given volume to find a missing length.
Written by Asad, Co-Founder of Teachably
Video walkthrough coming soon
The written lesson below covers everything you need in the meantime.
Volume and surface area of a sphere
The volume of a sphere is (4/3) × π × radius³, and its surface area is 4 × π × radius². A hemisphere is exactly half a sphere, so its volume is half the sphere volume, though its total surface area needs the curved half plus the flat circular face, not simply half the sphere surface area.
A sphere has radius 6 cm, so its volume is (4/3) × π × 6³ = 288π cm³, and its surface area is 4 × π × 6² = 144π cm².
Volume of a pyramid
The volume of any pyramid, whatever the shape of its base, is (1/3) × base area × perpendicular height. Find the area of the base first using the appropriate 2D area formula, then multiply by one third of the height.
A pyramid has a square base of side 6 cm and a perpendicular height of 8 cm. Its base area is 36 cm², so its volume is (1/3) × 36 × 8 = 96 cm³.
Volume and surface area of a cone
The volume of a cone is (1/3) × π × radius² × height, using the perpendicular height. The curved surface area is π × radius × slant height, where the slant height is the distance from the edge of the base to the apex, found using Pythagoras if only the perpendicular height is given. The total surface area adds the circular base, π × radius².
A cone has radius 3 cm and slant height 5 cm. Its curved surface area is π × 3 × 5 = 15π cm², and adding the base gives a total surface area of 15π + π × 3² = 24π cm².
Combined solids and working backwards
When two solids are joined, such as a cone topped with a hemisphere, find the volume of each part separately using the shared radius, then add them together. When a volume is given along with some dimensions, set up an equation using the correct formula and solve for the missing length.
An ice-cream is a cone of radius 3 cm and height 8 cm, topped with a hemisphere of the same radius. The cone volume is (1/3) × π × 3² × 8 = 24π cm³, and the hemisphere volume is (2/3) × π × 3³ = 18π cm³, giving a total of 42π cm³.
Worked Examples
Three exam-style questions, fully solved.
Find the volume of a sphere with radius 6 cm, leaving your answer in terms of π.
Easy- 1.Use the sphere volume formula: (4/3) × π × radius³ with radius 6
Answer: 288π cm³
A sphere has a volume of 36π cm³. Find the radius of the sphere.
Medium- 1.Set up an equation using the volume formula: 36π = (4/3) × π × r³
- 2.Multiply both sides by 3/4 and divide by π: r³ = 36 × 3 ÷ 4 = 27
- 3.Take the cube root to find the radius
Answer: r = 3 cm
Eight identical spherical marbles, each of radius 1.5 cm, are melted down and recast into a single cone of radius 3 cm. Find the height of the cone (assume no material is wasted).
Hard- 1.Find the volume of one marble: (4/3) × π × 1.5³ = 4.5π cm³
- 2.Find the total volume of all 8 marbles: 8 × 4.5π = 36π cm³
- 3.Set this equal to the cone volume: 36π = (1/3) × π × 3² × h
- 4.Solve for h: h = 36 × 3 ÷ 9
Answer: 12 cm
Avoid These
The most common mistakes students make.
Forgetting the (1/3) in the pyramid and cone volume formulae, effectively calculating the volume of a prism or cylinder instead.
Using the perpendicular height instead of the slant height (or the other way round) in the cone volume and curved surface area formulae, since volume needs the perpendicular height but curved surface area needs the slant height.
Halving the sphere surface area formula to find the total surface area of a hemisphere, instead of adding the flat circular face to half the curved surface area.
In melting and recasting problems, forgetting to find the total volume of all identical pieces before setting up the equation for the new shape.
When combining two solids, using the diameter as the radius for one part after it was correctly used as the radius for the other, especially when a cone and hemisphere share the same radius.
FAQ
Questions parents and students ask.
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