Question 32.

Assuming the dimensions of the brick are a, b and c and the diagonals are 3, 2 $\surd3$ and $\surd{15}$

Hence, $a^{2\ }+\ b^2$ = $3^2$  ......(1)

$b^{2\ }+\ c^2$ = $(2\sqrt{3})^2$ ......(2)

$c^{2\ }+\ a^2$ = $(\sqrt{15})^2$ ......(3)

Adding the three equations, 2($a^2+b^2+c^2$) = 9+12+15=36

=>$a^2+b^2+c^2$ = 18......(4)

Subtracting (1) from (4), we get $c^2$ = 9  =>c=3

Subtracting (2) from (4), we get $a^2$ = 6  =>a=$\sqrt{6}$

Subtracting (3) from (4), we get $b^2$ = 3  =>b=$\sqrt{3}$

The ratio of the length of the shortest edge of the brick to that of its longest edge is = $\ \frac{\ \sqrt{3}}{3}$ = $1 : \sqrt{3}$