Find the lengths of the diagonals

Geometry Level pending

The surface area of a cube is $1350$ . Find the sum of its space diagonal and face diagonal.

Note: In the figure above, $y$ is the face diagonal and $x$ is the space diagonal.

225\sqrt{2}+15\sqrt{3}

15

15(\sqrt{2}+\sqrt{3})

15\sqrt{2}+\sqrt{3}

1 solution

A Former Brilliant Member
Oct 31, 2017

$A=6a^2$

$1350=6a^2$

$a=15$

By pythagorean theorem, we have

$y=\sqrt{15^2+15^2}=\sqrt{450}=15\sqrt{2}$

By pythagorean theorem again, we have

$x=\sqrt{(15\sqrt{2})^2+15^2}=\sqrt{450+225}=\sqrt{675}=15\sqrt{3}$

The desired answer is $15\sqrt{2}+15\sqrt{3}=15(\sqrt{2}+\sqrt{3})$