Pyramids, cones and spheres

A pyramid’s perpendicular height controls volume

A pyramid has volume (1/3)Ah, where A is base area and h is perpendicular height to the base plane. A square base of side 6 cm and perpendicular height 4 cm gives volume 1/3 × 36 × 4 = 48 cm³.

For a right square pyramid with its apex above the base centre, each triangular face has base 6 cm and slant height square root of (4² + 3²) = 5 cm. Four faces have area 4(1/2 × 6 × 5) = 60 cm². Add the 36 cm² base for total 96 cm². The slant height is not the vertical height.

A right circular cone has two different heights too

For radius r, perpendicular height h and slant height l = square root of (r² + h²), cone volume is (1/3)π r²h. Its curved surface area is π rl; add π r² for a closed base.

With r = 3 cm and h = 4 cm, l = 5 cm. Volume is 12π cm³, curved area 15π cm² and total area 24π cm². A cone-shaped party hat normally omits the base, so it uses curved area rather than closed total area.

A sphere has no flat faces

A sphere of radius r has volume (4/3)π r³ and surface area 4π r². A hemisphere has half the sphere’s volume and curved area, but its closed total area also includes its circular base: 3π r².

For r = 3 cm, hemisphere volume is 18π cm³, curved area 18π cm² and closed total area 27π cm². State which surface the context includes.

Check your understanding 1 Choose the height for the purpose

A right cone has radius 5 cm and perpendicular height 12 cm. Find its slant height, volume and curved surface area.

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The slant height is 13 cm by Pythagoras. Volume is (1/3)π(25)(12) = 100π cm³. Curved area is π(5)(13) = 65π cm². Using 13 in the volume formula would use the wrong height.

Composite surfaces follow the exposed boundary

A cone and hemisphere of the same radius are joined along their flat circular bases. Add their volumes. For external area, add the cone’s curved area and hemisphere’s curved area; neither shared circular base remains exposed. Draw the join before deciding which formula terms to include.