Pyramid cut by a parallel plane

Problem

A right pyramid $P$ with square base of edge $4$ and height $3$ is cut by a plane parallel to the base at distance $x$ from it; $P'$ is the part above the cut. a) For $x=1{,}8$, compute the volume and total surface of $P$ and $P'$. b) Find the radius of the sphere inscribed in $P$. c) Find $x$ so that pyramid $P'$ and the frustum have volumes one triple the other.

Solution

a) $V_P=\dfrac13\cdot4^{2}\cdot3=16$. The face apothem is $\sqrt{2^{2}+3^{2}}=\sqrt{13}$, so $S_P=16+4\cdot\dfrac{4\sqrt{13}}{2}=16+8\sqrt{13}\approx44{,}84$. Pyramid $P'$ is similar with ratio $k=\dfrac{3-1{,}8}{3}=0{,}4$: $V_{P'}=k^{3}V_P\approx1{,}02$, $S_{P'}=k^{2}S_P\approx7{,}17$. b) For a pyramid the inscribed sphere has radius $r=\dfrac{3V}{S_{\text{tot}}}=\dfrac{48}{16+8\sqrt{13}}\approx1{,}07$. c) $V_{P'}=k^{3}V_P$ with $k=\dfrac{3-x}{3}$. If the frustum is triple $P'$: $16-V_{P'}=3V_{P'}\Rightarrow V_{P'}=4\Rightarrow k^{3}=\tfrac14\Rightarrow x=3\left(1-\sqrt[3]{1/4}\right)\approx1{,}11$. (In the other case $x\approx0{,}27$.) $\boxed{\,V_P=16,\ S_P=16+8\sqrt{13}\approx44{,}84,\ r\approx1{,}07,\ x\approx1{,}11\,}$