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.

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Complete 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.

B. StartFraction pi Over 4 EndFraction the volume of the pyramid or StartFraction pi Over 4 EndFraction (StartFraction (2 r) squared (h) Over 3 EndFraction) or One-thirdπr²h = 1/3πr²h

Step-by-step explanation:

We have two geometric shapes in the question.

a) A cone and b) a square pyramid

The cone has a height of h and a radius of r. The pyramid has a base length of 2 r.

The volume of a cone =1/3πr²h

Where πr² = Area of the circle at the base of the cone

Hence, Volume of a cone = 1/3 × Area of the circular base of a cone × Height

The volume of a square pyramid = 1/3a²h

Where a² = Area of the square base of the pyramid

Hence, Volume of a square pyramid = 1/3 × Area of the square base of a pyramid × height(h)

Base area of a cone / Base area of a square pyramid = π/4

Base area of a circle = Base area of a pyramid × π/4

Volume of a cone = 1/3πr²h

Volume of a cone = 1/3 × Base area of a square pyramid × π/4 × h

Note that:

Volume of a square pyramid = 1/3a²h

= 1/3 × Base area of a square pyramid × height

Hence,

Volume of a cone = Volume of a square pyramid × π/4

= StartFraction pi Over 4 EndFraction the volume of the pyramid

Or

Where a = base length = 2r

Volume of the square pyramid = 1/3 × 2r² × h = 1/3 × 4r²h

Volume of a cone = Volume of a square pyramid × π/4

Substituting = 1/3 × 4 × r²× h × π/4

Volume of a cone = 1/3 πr²h

Or

Volume of a cone = Volume of a square pyramid × π/4

Volume of a square pyramid when base length is 2r = 1/3 × (2r)² × h = (2r²)h/3

Substituting (2r²)h/3 for volume of a square pyramid in volume of a cone , we have:

Volume of a cone = π/4 × 2r²h/3

=

StartFraction pi Over 4 EndFractionStartFraction pi Over 4 EndFraction (StartFraction (2 r) squared

Therefore, Option B is correct

B

Step-by-step explanation: