Question

A pyramid of height 15 ft is to be divided into three solids by passing

planes parallel to the base. Find the height of each solid if their volumes

are equal.

Answer 1

The volume of whole this pyramide is:

$V=\frac{1}{3}Sh=\frac{1}{3}⋅S⋅15=5S$

According to the conditions of the problem

$V_1=V_2=V_3$,

therefore

$V=V_1+V_2+V_3$

$V=3V_1$

$5S=3V_1$

$V_1=\frac{5S}{3}$

The volume of the top pyramide is:

$V_1=\frac{1}{3}S_1h_1$, therefore

$\frac{5S}{3}=\frac{1}{3}S_1h_1$

$5S=S_1h_1$

$h_1=\frac{5S}{S_1}$ - it is the height of the top pyramide

Then

$V_1+V_2=\frac{1}{3}S_2(h_1+h_2)$

$2V_1=\frac{1}{3}S_2(h_1+h_2)$

$2V_1=\frac{1}{3}S_2(\frac{5S}{S_1}+h_2)$ $/⋅3$

$6V_1=S_2(\frac{5S}{S_1}+h_2)$

$\frac{5S}{S_1}+h_2=\frac{6V_1}{S_2}$

$h_2=\frac{6V_1}{S_2}-\frac{5S}{S_1}$

$h_2=\frac{6⋅\frac{5S}{3}}{S_2}-\frac{5S}{S_1}$

$h_2=\frac{10S}{S_2}-\frac{5S}{S_1}$ - it is the height of the middle solide

$h_3=15-h_1-h_2=15-\frac{5S}{S_1}-(\frac{10S}{S_2}-\frac{5S}{S_1})=$

$=15-\frac{5S}{S_1}-\frac{10S}{S_2}+\frac{5S}{S_1}=15-\frac{10S}{S_2}$ - it is the height of the under solid

Solution: $\frac{5S}{S_1}$ ; $\frac{10S}{S_2}-\frac{5S}{S_1}$ ; $15-\frac{10S}{S_2}$ .