Mensuration II

The following results are very important to solve various mensuration problems.

1.  The largest possible sphere that can be chiseled out from a cube of side "a" cm. 

 Diagonal of the sphere is a, so radius = a/2.
Remaining empty space in the cube = a3−πa36$a^3-{\displaystyle \frac{\pi a^3}{6}}$
${\displaystyle \frac{\mathrm{<span\; style="color:\; blue;\; font-size:\; 13.65px;\; line-height:\; 27.3px;\; white-space:\; normal;">2.\; The\; largest\; possible\; cube\; that\; can\; be\; chiseled\; out\; from\; a\; sphere\; of\; radius\; "a"\; cm</span>}}{}}$

Here OA = radius of the sphere. So diameter of the sphere = 2a.

Let the side of the square = x$x$, then the diagonal of the cube = 3√x$\sqrt{3}x$

⇒3√x=2a$\Rightarrow \sqrt{3}x=2a$ ⇒x=2a3√$\Rightarrow x={\displaystyle \frac{2a}{\sqrt{3}}}$

${\displaystyle \frac{\mathrm{}}{}}$

Therefore side of the square = 2a3√${\displaystyle \frac{2a}{\sqrt{3}}}$ 

3.  The largest possible cube that can be chiseled out from a hemisphere of radius 'a' cm. 

Sol:

Given, the radius of the hemi sphere AC = a$a$.  Let the side of the cube is x$x$.
From the above diagram, BE2+ED2=BD2$B{E}^2+E{D}^2=B{D}^2$
⇒x2+x2=BD2$\Rightarrow x^2+x^2=B{D}^2$
⇒BD=2√x$\Rightarrow BD=\sqrt{2}x$
⇒BC=2√x2=x2√$\Rightarrow BC={\displaystyle \frac{\sqrt{2}x}{2}}={\displaystyle \frac{x}{\sqrt{2}}}$
From Δ$\mathrm{\Delta }$ABC, AC2=AB2+BC2$A{C}^2=A{B}^2+B{C}^2$
⇒a2=x2+(x2√)2$\Rightarrow a^2=x^2+{\left({\displaystyle \frac{x}{\sqrt{2}}}\right)}^2$
⇒a2=3x22$\Rightarrow a^2={\displaystyle \frac{3x^2}{2}}$
⇒x=23−−√a$\Rightarrow x=\sqrt{{\displaystyle \frac{2}{3}}}a$
The edge of the cube = a23−−√