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Let the side of a brick in a cuboid shape be $a, b,$ and $c$. Where $a>b>c.$

We can draw the diagram:

Given that, ratio of the diagonals $ = 3 : 2 \sqrt{3} : \sqrt{15}$

$ \Rightarrow \sqrt{b^{2} + c^{2}} : \sqrt{ c^{2} + a^{2}} : \sqrt{a^{2} + b^{2}} = 3 : 2 \sqrt{3} : \sqrt{15} \quad [\because \text{Using Pythagoras’ theorem}]$

On squaring both sides, we get

- $ b^{2} + c^{2} = 9 \quad \longrightarrow (1)$
- $ c^{2} + a^{2} = 12 \quad \longrightarrow (2)$
- $ a^{2} + b^{2} = 15 \quad \longrightarrow (3)$

Subtract the equation $(2),$ from the equation $(1),$ we get

$\begin{array}{c} b^{2} + c^{2} = 9 \\ a^{2} + c^{2} = 12 \\ \hline b^{2} – a^{2} = -3 \end{array}$

$ \Rightarrow a^{2} – b^{2} = 3 \quad \longrightarrow (4) $

Adding the equation $(3),$ and $(4),$ we get

$\require{cancel}\begin{array}{c} a^{2} + \cancel{b^{2}} = 15 \\ a^{2} \; – \; \cancel{b^{2}} = 3 \\ \hline 2 a^{2} = 18 \end{array}$

$ \Rightarrow a^{2} = 9$

$ \Rightarrow \boxed{a=3}$

Now, from the equation $(2),$ we get

$ c^{2} + (3)^{2} = 12$

$\Rightarrow c^{2} = 12-9$

$ \Rightarrow c^{2} = 3$

$\Rightarrow \boxed{c = \sqrt{3}}$

$ \therefore$ The ratio of the length of the shortest edge of the brick to that of its longest edge of the brick $= \frac{c}{a} = \frac{\sqrt{3}}{3} = \frac{\sqrt{3}}{3} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{3}{3 \sqrt{3}} = \frac{1} { \sqrt{3}}.$

Correct Answer $: \text{C}$