Question

The base of right pyramid, $h$ cm height is a square. It is divided into two parts by a plane parallel to the base so that volumes of the two parts are equal. If the distance of the plane form the vertex is $\frac{h}{\sqrt[a]{ab}}$ cm, then find the value of $(a-b{)}^2$.

Answer 1

Volume of square base pyramid$=\frac{1}{3}\times B_A\times h$

$B_A=$Area of base

(Image)

By symmetry,

$\frac{h^1}{h}=\frac{l^1}{l}$

Acc. to question,

$\frac{1}{3}{B}_a\times h=(\frac{1}{3}\times B_a^1\times h^1)\times 2$

$\Rightarrow \frac{1}{3}\times l^{2}h=\frac{2}{3}{\left(l^1\right)}^2{h}^1$

$\Rightarrow \frac{1}{2}={\left(\frac{l^1}{l}\right)}^2\times \frac{h^1}{h}$

${\left(\frac{h^1}{h}\right)}^3=\frac{1}{2}$

$h^1=\frac{h}{\sqrt[3]{32}}$

$a=3$

$b=2$

${(a-b)}^2=1^2=1$