Sides of a triangle are in the ratio $3 : 4 : 5$ and its area is $54 c m^{2}$ . Find the sides of the triangle.

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Solution

$\begin{matrix} L e t s i d e s b e 3 x, 4 x a n d 5 x S = \frac{3 x + 4 x + 5 x}{2} S = \frac{12 x}{2} = 6 x A r e a o f Δ = \sqrt{6 x (3 x) (2 x) x} 54 = \sqrt{36 x^{4}} 56 = 6 x^{2} x^{2} = 9 x = 3 u n i t H e n c e, s i d e s o f t r i a n g l e a r e 3 x = 9 u n i t 4 x = 12 u n i t 5 x = 15 u n i t \end{matrix}$

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