The sides of a triangle having area 7776 cm^ | vedclass

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(D) Let the sides of the triangle be $3x, 4x,$ and $5x$.Since the ratio $3:4:5$ satisfies the Pythagorean theorem $(3^2 + 4^2 = 5^2)$,the triangle is

Vedclass · Accepted Answer

$432$

Vedclass · Answer

(D) Let the sides of the triangle be $3x, 4x,$ and $5x$.Since the ratio $3:4:5$ satisfies the Pythagorean theorem $(3^2 + 4^2 = 5^2)$,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 $= \frac{1}{2} 	imes 3x 	imes 4x = 6x^2$.Given that the area is $7776 \ cm^2$,we have $6x^2 = 7776$.$x^2 = \frac{7776}{6} = 1296$.$x = \sqrt{1296} = 36$.The perimeter of the triangle is the sum of its sides: $3x + 4x + 5x = 12x$.Perimeter $= 12 	imes 36 = 432 \ cm$.

The sides of a triangle having area $7776 \ cm^2$ are in the ratio $3:4:5$. The perimeter of the triangle is ........ $cm$.

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