The sides of a triangle are 50 , cm,78 , cm,a | vedclass

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(B) The semi-perimeter of the triangle $(s)$ is calculated as:$s = \frac{50 + 78 + 112}{2} = \frac{240}{2} = 120 \, cm$.Using Heron's formula,the area

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$30$

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(B) The semi-perimeter of the triangle $(s)$ is calculated as:$s = \frac{50 + 78 + 112}{2} = \frac{240}{2} = 120 \, cm$.Using Heron's formula,the area of the triangle is:$	ext{Area} = \sqrt{s(s-a)(s-b)(s-c)}$$= \sqrt{120(120-50)(120-78)(120-112)}$$= \sqrt{120 	imes 70 	imes 42 	imes 8} = \sqrt{2822400} = 1680 \, cm^2$.The altitude is smallest when the base is the largest side.Let the base $b = 112 \, cm$ and the smallest altitude be $h$.$	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$$1680 = \frac{1}{2} 	imes 112 	imes h$$1680 = 56 	imes h$$h = \frac{1680}{56} = 30 \, cm$.

The sides of a triangle are $50 \, cm$,$78 \, cm$,and $112 \, cm$. The smallest altitude is $........... \, cm$.

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