The sides of a triangle are 35 text{ cm},54 t | vedclass

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(D) The sides of the triangle are $a = 35 	ext{ cm}$,$b = 54 	ext{ cm}$,and $c = 61 	ext{ cm}$.First,calculate the semi-perimeter $s$:$s = \frac{35

Vedclass · Accepted Answer

$24 \sqrt{5} 	ext{ cm}$

Vedclass · Answer

(D) The sides of the triangle are $a = 35 	ext{ cm}$,$b = 54 	ext{ cm}$,and $c = 61 	ext{ cm}$.First,calculate the semi-perimeter $s$:$s = \frac{35 + 54 + 61}{2} = \frac{150}{2} = 75 	ext{ cm}$.Using Heron's formula,the area of the triangle is:$	ext{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{75(75-35)(75-54)(75-61)}$$= \sqrt{75 	imes 40 	imes 21 	imes 14} = \sqrt{(25 	imes 3) 	imes (8 	imes 5) 	imes (3 	imes 7) 	imes (2 	imes 7)}$$= \sqrt{25 	imes 3^2 	imes 7^2 	imes 16 	imes 5} = 5 	imes 3 	imes 7 	imes 4 \sqrt{5} = 420 \sqrt{5} 	ext{ cm}^2$.The longest altitude corresponds to the smallest side of the triangle.Since the smallest side is $35 	ext{ cm}$,the longest altitude $h$ is given by:$	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} \implies 420 \sqrt{5} = \frac{1}{2} 	imes 35 	imes h$.$h = \frac{420 \sqrt{5} 	imes 2}{35} = 12 	imes 2 \sqrt{5} = 24 \sqrt{5} 	ext{ cm}$.

The sides of a triangle are $35 \text{ cm}$,$54 \text{ cm}$,and $61 \text{ cm}$,respectively. The length of its longest altitude is:

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