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 $\frac{pi r^{2}}{4r^{2}}$ or $\frac{pi}{4}$. since the area of the circle is $\frac{pi}{4}$ the area of the square, the volume of the cone equals $\frac{pi}{4}$ the volume of the pyramid or $\frac{pi}{4}left\frac{(2r)(h)}{3}
ight$ or $\frac{1}{6}pi rh$. $\frac{pi}{4}$ the volume of the pyramid or $\frac{pi}{4}left\frac{(2r)^{2}(h)}{3}
ight$ or $\frac{1}{3}pi r^{2}h$. $\frac{pi}{2}$ the volume of the pyramid or $\frac{pi}{2}left\frac{(2r)^{2}(h)}{3}
ight$ or $\frac{2}{3}pi r^{2}h$. $\frac{pi}{2}$ the volume of the pyramid or $\frac{pi}{4}left\frac{(2r)^{2}(h)}{3}
ight$ or $\frac{1}{3}pi r^{2}h$. cross section

Explanation:

Step1: Recall volume formulas

Volume of a cone $V_{cone}=\frac{1}{3}\pi r^{2}h$, volume of a square - pyramid $V_{pyramid}=\frac{1}{3}s^{2}h$. Here, the side - length of the base of the square - pyramid $s = 2r$. So $V_{pyramid}=\frac{1}{3}(2r)^{2}h=\frac{4}{3}r^{2}h$.

Step2: Use the cross - section ratio

We know that the ratio of the area of the cross - section of the circle to the area of the cross - section of the square is $\frac{\pi}{4}$. Since the height of the cone and the pyramid is the same, the volume of the cone $V_{cone}$ and the volume of the pyramid $V_{pyramid}$ are related by the same ratio of the cross - sectional areas.
$V_{cone}=\frac{\pi}{4}V_{pyramid}$. Substitute $V_{pyramid}=\frac{(2r)^{2}h}{3}$ into the equation: $V_{cone}=\frac{\pi}{4}\times\frac{(2r)^{2}h}{3}=\frac{\pi}{4}\times\frac{4r^{2}h}{3}=\frac{1}{3}\pi r^{2}h$.

Answer:

$\frac{\pi}{4}$ the volume of the pyramid or $\frac{\pi}{4}(\frac{(2r)^{2}(h)}{3})$ or $\frac{1}{3}\pi r^{2}h$ (Second option)