The figure shows a solid frustum of height 16 cm cut from a right pyramid with square base. The horizontal top surface has an area of 5 cm^2 while the base area is 125 cm^2. The portion that is removed is $h$ cm high. Show that $h = 4$. Answer Calculate the volume of the frustum. Answer ... cm^3 Find the percentage of the right pyramid that has been removed. Answer ... % Two solid cones have the same height but the radius of cone A is 1.5 times of cone B. Given that the volume of cone B is 240 cm^3. Find the volume of cone A.

1. Set up the area ratio of the similar pyramids

The removed small pyramid and the original pyramid are similar.

Area of top square $=5\text{ cm}^2$ and area of base square $=125\text{ cm}^2$.

$\frac{\text{area of top square}}{\text{area of base square}}=\frac{5}{125}=\frac{1}{25}$

2. Convert area ratio to height ratio and solve for $h$

For similar pyramids,

$\frac{\text{small height}}{\text{original height}}=\sqrt{\frac{1}{25}}=\frac{1}{5}$

The small height is $h$ and the original height is $h+16$.

$\frac{h}{h+16}=\frac{1}{5}$

$5h=h+16$

$4h=16$

$h=4$

1. Find the height of the original right pyramid

From part (a)(i), the removed small pyramid has height $h=4$ cm.

Height of original pyramid
$=4+16=20\text{ cm}$

2. Calculate the volumes of the original pyramid and the removed small pyramid

Volume of a pyramid $=\dfrac{1}{3}\times \text{base area}\times \text{height}$.

Original pyramid:
$V_{\text{large}}=\frac{1}{3}\times 125\times 20=\frac{2500}{3}$

Removed pyramid:
$V_{\text{small}}=\frac{1}{3}\times 5\times 4=\frac{20}{3}$

3. Subtract to find the volume of the frustum

Volume of frustum
$=V_{\text{large}}-V_{\text{small}}$
$=\frac{2500}{3}-\frac{20}{3}$
$=\frac{2480}{3}=826.666\ldots$
$\approx 827\text{ cm}^3$

1. Identify the relevant volumes from earlier work

From 7(a)(ii),

height of original pyramid $= h+16=4+16=20\text{ cm}$.

Volume of original pyramid:
$\frac13 \times 125 \times 20=\frac{2500}{3}\text{ cm}^3$

Volume of removed small pyramid:
$\frac13 \times 5 \times 4=\frac{20}{3}\text{ cm}^3$

2. Calculate the percentage of the original pyramid that was removed

Percentage removed
$=\frac{\text{volume removed}}{\text{volume of original pyramid}}\times 100\%$

$=\frac{\frac{20}{3}}{\frac{2500}{3}}\times 100\%$

$=\frac{20}{2500}\times 100\%=0.8\%$

1. Relate the volumes using the radius scale factor

For a cone,
$V=\frac{1}{3}\pi r^2h.$
The cones have the same height, so
$\frac{V_A}{V_B}=\left(\frac{r_A}{r_B}\right)^2.$
Given $r_A=1.5r_B$,
$\frac{V_A}{V_B}=1.5^2=2.25.$

2. Calculate the volume of cone A

$V_A=2.25\times 240=540.$