The sides of a triangle are in the ratio 3:4: | vedclass

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(B) Let the sides of the triangle be $3x$,$4x$,and $5x$.Given that the perimeter is $36 	ext{ cm}$,we have:$3x + 4x + 5x = 36$$12x = 36$$x = 3$Theref

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$54$

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(B) Let the sides of the triangle be $3x$,$4x$,and $5x$.Given that the perimeter is $36 	ext{ cm}$,we have:$3x + 4x + 5x = 36$$12x = 36$$x = 3$Therefore,the sides are:$a = 3(3) = 9 	ext{ cm}$$b = 4(3) = 12 	ext{ cm}$$c = 5(3) = 15 	ext{ cm}$The semi-perimeter $s$ is given by:$s = \frac{a+b+c}{2} = \frac{36}{2} = 18 	ext{ cm}$Using Heron's formula,the area of the triangle is:$	ext{Area} = \sqrt{s(s-a)(s-b)(s-c)}$$	ext{Area} = \sqrt{18(18-9)(18-12)(18-15)}$$	ext{Area} = \sqrt{18 	imes 9 	imes 6 	imes 3}$$	ext{Area} = \sqrt{2916} = 54 	ext{ cm}^2$Alternatively,since $9^2 + 12^2 = 81 + 144 = 225 = 15^2$,it is a right-angled triangle.$	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 9 	imes 12 = 54 	ext{ cm}^2$.

The sides of a triangle are in the ratio $3:4:5$. If its perimeter is $36 \text{ cm}$,then the area of the triangle is ...... $\text{cm}^2$.

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