Question

The altitude of the frustum of a regular rectangular pyramid is $5m$ the volume is $140cu.m.$ and the upper base is $3m$ by $4m$. What are the dimensions of the lower base in $m$?

• A. $9\times 10$
• B. $6\times 8$
• C. $4.5\times 6$
• D. $7.5\times 10$

Answer 1

The correct option is B $6\times 8$

Given : Height of pyramid$\left(h\right)=5m$, Volume$\left(V\right)=140cu.m$

Edge of upper base are $length\left(L\right)=3m,breadth\left(B\right)=4m$

And let $l,h$ be the length and height of lower base

$\therefore$ Area of upper base$\left(A_1\right)=L\times B=3\times 4=12m^2$

And Area of lower base$A_2=l\times b$

Volume of frustum$\left(V\right)=\frac{h}{3}(A_1+A_2+\sqrt{A_1\times A_2})$

$⟹$$140=\frac{5}{3}(12+A_2+\sqrt{12\times A_2})$

$⟹84=A_2+12+\sqrt{12}\sqrt{A_2}$

$⟹A_2+\sqrt{12}\sqrt{A_2}-72=0$

$⟹{(\sqrt{A_2}+\frac{\sqrt{12}}{2})}^2-\frac{12}{4}-72=0$

$⟹{(\sqrt{A_2}+\frac{\sqrt{12}}{2})}^2=75$

$⟹(\sqrt{A_2}+\frac{\sqrt{12}}{2})=\sqrt{75}$

$⟹\sqrt{A_2}=6.928$

$\therefore A_2=47.9971=48m^2=6\times 8$

Hence, dimensions of lower base is $6\times 8$.