The ratios of the sides of a triangle ABC are | vedclass

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(C) Let the sides of the triangle be $5x$,$12x$,and $13x$.Since $5^2 + 12^2 = 25 + 144 = 169 = 13^2$,the triangle is a right-angled triangle.The area

Vedclass · Accepted Answer

$15, 36, 39$

Vedclass · Answer

(C) Let the sides of the triangle be $5x$,$12x$,and $13x$.Since $5^2 + 12^2 = 25 + 144 = 169 = 13^2$,the triangle is a right-angled triangle.The area of a right-angled triangle is given by $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$.Area $= \frac{1}{2} 	imes 5x 	imes 12x = 30x^2$.Given that the area is $270$,we have $30x^2 = 270$.$x^2 = 9$,which implies $x = 3$.The sides are $5(3) = 15$,$12(3) = 36$,and $13(3) = 39$.Thus,the sides are $15, 36, 39$.

The ratios of the sides of a triangle $ABC$ are $5:12:13$ and its area is $270$. Then the sides of the triangle are:

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