If the rectangular faces of a brick have their diagonals in the ratio $$3 : 2 \surd3 : \surd{15}$$, then the ratio of the length of the shortest edge of the brick to that of its longest edge is

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 = $$\ \frac{\ \sqrt{3}}{3}$$ = $$1 : \sqrt{3}$$