Numerical answer
A right circular cone has base radius 9 cm and height 21 cm. It is cut by a plane parallel to its base at a height of 14 cm above the base. Find the volume (in cm³) of the frustum that remains below the cut. (Take π = 22/7.)
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Answer: 1716
Volume of the whole cone = (1/3)π × 9² × 21 = 567π.
The cut is 14 cm above the base, so the small cone on top has height 21 − 14 = 7 cm — a scale factor of 7/21 = 1/3 of the whole cone, with radius 9/3 = 3 cm. Its volume = (1/3)³ × 567π = 21π.
Frustum = 567π − 21π = 546π = 546 × 22/7 = 1,716 cm³.
Tip: a cut at fraction k of the height from the apex leaves a top cone of volume k³ of the whole; here k = 1/3 so the top cone is 1/27 of the total volume.
The cut is 14 cm above the base, so the small cone on top has height 21 − 14 = 7 cm — a scale factor of 7/21 = 1/3 of the whole cone, with radius 9/3 = 3 cm. Its volume = (1/3)³ × 567π = 21π.
Frustum = 567π − 21π = 546π = 546 × 22/7 = 1,716 cm³.
Tip: a cut at fraction k of the height from the apex leaves a top cone of volume k³ of the whole; here k = 1/3 so the top cone is 1/27 of the total volume.
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