The sides of a right-angled triangle forming | vedclass

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(A) Let the perpendicular sides of the right-angled triangle be $5x$ and $12x$.According to the question,the area of the triangle is given by:$\frac{1

Vedclass · Accepted Answer

$39$

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(A) Let the perpendicular sides of the right-angled triangle be $5x$ and $12x$.According to the question,the area of the triangle is given by:$\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = 270 \, cm^2$$\frac{1}{2} 	imes 5x 	imes 12x = 270$$30x^2 = 270$$x^2 = \frac{270}{30} = 9$$x = 3$Now,the sides are $5 	imes 3 = 15 \, cm$ and $12 	imes 3 = 36 \, cm$.The hypotenuse is given by the Pythagorean theorem:$	ext{Hypotenuse} = \sqrt{(15)^2 + (36)^2} = \sqrt{225 + 1296} = \sqrt{1521} = 39 \, cm$.Alternatively,using the ratio $5:12:13$,the hypotenuse is $13x = 13 	imes 3 = 39 \, cm$.

The sides of a right-angled triangle forming the right angle are in the ratio $5: 12$. If the area of the triangle is $270 \, cm^2$,then the length of the hypotenuse is .......... $cm$.

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