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

A. 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.

B. 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.

C. StartFraction pi Over 2 EndFraction the volume of the pyramid or StartFraction pi Over 2 EndFraction or Two-thirdsπr2h.

D. StartFraction pi Over 2 EndFraction the volume of the pyramid or StartFraction pi Over 4 EndFraction or One-thirdπr2h.

Answer 1

Answer:

B

Step-by-step explanation:

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Answer 2

Answer:

Option B

Step-by-step explanation:

we know that

The volume of the cone is equal to

$V_c=\frac{1}{3}B_c(h)$

where

Bc is the area of the circle of the base of the cone

The volume of the square pyramid is equal to

$V_p=\frac{1}{3}B_p(h)$

where

Bp is the area of the square base of the pyramid

we know that

$\frac{B_c}{B_p}=\frac{\pi}{4}$

$B_c=\frac{\pi}{4}(B_p)$

substitute in the formula of volume of the cone

$V_c=\frac{1}{3}B_c(h)$

$V_c=\frac{1}{3}(\frac{\pi}{4}(B_p))(h)$

Remember that

$V_p=\frac{1}{3}B_p(h)$

substitute

$V_c=(\frac{\pi}{4})V_p$ ----> StartFraction pi Over 4 EndFraction the volume of the pyramid

or

$V_c=(\frac{\pi}{4})(\frac{(2r)^2h}{3})$ ----> StartFraction pi Over 4 EndFraction (StartFraction (2 r) squared (h) Over 3 EndFraction)

or

$V_c=\frac{1}{3}\pi r^{2} h$ ----> One-thirdπr^2h