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 equalsA. 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.
Accepted Solution
A:
Answer:Option BStep-by-step explanation:we know thatThe volume of the cone is equal to[tex]V_c=\frac{1}{3}B_c(h)[/tex]whereBc is the area of the circle of the base of the coneThe volume of the square pyramid is equal to[tex]V_p=\frac{1}{3}B_p(h)[/tex]whereBp is the area of the square base of the pyramidwe know that[tex]\frac{B_c}{B_p}=\frac{\pi}{4}[/tex][tex]B_c=\frac{\pi}{4}(B_p)[/tex] substitute in the formula of volume of the cone[tex]V_c=\frac{1}{3}B_c(h)[/tex][tex]V_c=\frac{1}{3}(\frac{\pi}{4}(B_p))(h)[/tex]Remember that[tex]V_p=\frac{1}{3}B_p(h)[/tex]substitute[tex]V_c=(\frac{\pi}{4})V_p[/tex] ----> StartFraction pi Over 4 EndFraction the volume of the pyramidor[tex]V_c=(\frac{\pi}{4})(\frac{(2r)^2h}{3})[/tex] ----> StartFraction pi Over 4 EndFraction (StartFraction (2 r) squared (h) Over 3 EndFraction)or[tex]V_c=\frac{1}{3}\pi r^{2} h[/tex] ----> One-thirdπr^2h