Question

A cone fits inside a square pyramid as shown. For every cross section, the ratio of the area of the circle to the area of the square is StartFraction pi r squared Over 4 r squared EndFraction or StartFraction pi Over 4 EndFraction.

A cone is inside of a pyramid with a square base. The cone has a height of h and a radius of r. The pyramid has a base length of 2 r.

Since the area of the circle is StartFraction pi Over 4 EndFraction the area of the square, the volume of the cone equals

StartFraction pi Over 4 EndFraction the volume of the pyramid or StartFraction pi Over 4 EndFractionStartFraction pi Over 4 EndFraction (StartFraction (2 r) (h) Over 3 EndFraction) or One-sixthπrh.
StartFraction pi Over 4 EndFraction the volume of the pyramid or StartFraction pi Over 4 EndFractionStartFraction pi Over 4 EndFraction (StartFraction (2 r) squared (h) Over 3 EndFraction) or One-thirdπr2h.
StartFraction pi Over 2 EndFraction the volume of the pyramid or StartFraction pi Over 2 EndFraction or Two-thirdsπr2h.
StartFraction pi Over 2 EndFraction the volume of the pyramid or StartFraction pi Over 4 EndFraction or One-thirdπr2h.

Answer 1

Because every cross-section's circle area is (π/4) times the square area, the cone's volume is (π/4) times the pyramid's volume.

Volume of the pyramid = (1/3)·(base area)·h = (1/3)·(2r)^2·h = (4/3)r^2h.

So volume of the cone = (π/4)·(4/3)r^2h = (1/3)πr^2h.

Thus the correct result is (1/3)πr^2h (the option that gives one-third π r^2 h).