Question

Consider a triangular prism formed by the coordinate planes and planes $P_1:z=4$ and $P_2=x$ $+y=2$. If $S$ denotes its surface area, $V$ denotes its volume and $d$ denotes the shortest distance between its edge lying in $xy$-plane and $z$-axis, then find the value of $\frac{2S-5V}{8d}$.

Answer 1

Answer: 2

$OM=d=\frac{p+0-21}{\sqrt{2}}=\sqrt{2}$

$\text{ Volume }=\frac{1}{2}\times 2\times 2\times 4=8\text{ cubic units }$
$\text{ Surface area }=2\left(\frac{1}{2}\cdot 2\cdot 2\right)+4\cdot 2\sqrt{2}+4\cdot 2\cdot 2=20+8\sqrt{2}$
$\therefore \frac{2S-5V}{8d}=2$