Surface area and volume of right circular cones

Learning outcome

Calculate the surface area and volume of a right circular cone using the correct perpendicular or slant height.

Concepts and assumptions

A right circular cone has radius r and its apex directly above the base centre. Perpendicular height h joins centre and apex; slant height s joins apex and rim along the side. A right triangle gives s² = r² + h², so s = √(r² + h²).

Opening the curved surface produces a circular sector with radius s and arc length 2 × π × r. Its area is half the radius times the arc length. Therefore curved surface area, CSA = π × r × s, also called lateral area.

A closed cone includes its base: total surface area, TSA = π × r × s + π × r². A floorless tent or open conical vessel has no disk across its opening.

Volume = π × r² × h ÷ 3. Circular slices shrink towards the apex; geometry gives one-third of the cylinder with the same base and perpendicular height, not slant height.

Use π = 22/7 as the prescribed approximation. Dimensions are exact; capacity uses internal dimensions. Ignore thickness, seams and waste. Keep the stated centimetre or metre units consistent. No additional rounding is needed.

Worked examples

Example 1 — Closed cone. Radius is 7 cm and perpendicular height is 24 cm. Slant height = √(49 + 576) = √625 = 25 cm. CSA = (22/7) × 7 × 25 = 550 cm². TSA = 550 + 154 = 704 cm². Volume = 154 × 24 ÷ 3 = 1232 cm³.

Example 2 — Tent covering. A floorless conical tent has radius 3.5 m and perpendicular height 12 m. Its curved side is completely covered. Slant height = √(12.25 + 144) = 12.5 m. Canvas area = (22/7) × 3.5 × 12.5 = 137.5 m². At ₹48 per m², cost = 137.5 × 48 = ₹6600.

Example 3 — Reverse calculation. A right cone has diameter 42 cm and volume 12936 cm³. Radius = 21 cm; base area = (22/7) × 21² = 1386 cm². Height = 3 × 12936 ÷ 1386 = 28 cm. Slant height = √(441 + 784) = 35 cm. Therefore CSA = (22/7) × 21 × 35 = 2310 cm².

Common mistakes

Use s for curved area but h for volume. Do not add a floor to a floorless tent. A truncated cone is a different solid; these formulas assume the apex is present.

Practice questions

1. A cone has radius 21 cm and perpendicular height 20 cm. Find slant height and volume.
2. A closed cone has radius 7 cm and slant height 13 cm. Find total surface area.
3. A floorless conical tent has radius 10.5 m and perpendicular height 14 m. Find canvas area and cost at ₹40 per m².
4. An open conical vessel has internal radius 14 cm and capacity 4312 cm³. Find its perpendicular depth.

Worked solutions

1. Slant height = √(441 + 400) = 29 cm. Volume = (22/7) × 441 × 20 ÷ 3 = 9240 cm³.
2. CSA = (22/7) × 7 × 13 = 286 cm². Add the base: TSA = 286 + 154 = 440 cm².
3. Slant height = √(110.25 + 196) = 17.5 m. Area = (22/7) × 10.5 × 17.5 = 577.5 m². Cost = 577.5 × 40 = ₹23100.
4. Base area = (22/7) × 14² = 616 cm². Depth = 3 × 4312 ÷ 616 = 21 cm.