Question

Consider the cube given below:
$NR,MS$ and $PT$ are the diagonals of the cube whose volume is $1728c{m}^3$. If a triangle is constructed whose three sides are same as the measure of the sides $NR,MS$ and $PT$, then what will be the radius of the circle circumscribing that triangle?

• A. $12cm$
• B. $12\sqrt{3}cm$
• C. $\frac{24}{\sqrt{5}}cm$
• D. $24cm$
• E. $noneofthese$

Answer 1

The correct option is C $\frac{24}{\sqrt{5}}cm$

Volume of cube$=1728㎠$

Side of cube $a=\sqrt[3]{31728}$

$=12$

$\therefore NR=\sqrt{2}.12=12\sqrt{2}=MS$

$PT=\sqrt{3}.12=12\sqrt{3}$

$\therefore$Sides of $△$ are $12\sqrt{2},12\sqrt{2},12\sqrt{3}㎝$

$h=\sqrt{{\left(12\sqrt{2}\right)}^2-\frac{{\left(12\sqrt{3}\right)}^2}{4}}$

$=\sqrt{288-108}$

$=\sqrt{180}$

Area$=\frac{1}{2}\times \sqrt{180}\times 12\sqrt{3}=6\sqrt{180\times 3}$

$=6\sqrt{4\times 3\times 5\times 9}$

$=6\times 3\times 2\sqrt{15}$

$=36\sqrt{15}$

$R=\frac{abc}{4△}=\frac{12\sqrt{2}\times 12\sqrt{2}\times 12\sqrt{3}}{4\times 36\times \sqrt{15}}=\frac{24}{\sqrt{5}}$