The edges of a triangular board are 6 , cm,8 | vedclass

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(B) The sides of the triangle are $a = 6 \, cm$,$b = 8 \, cm$,and $c = 10 \, cm$.First,calculate the semi-perimeter $s$:$s = \frac{a + b + c}{2} = \fr

Vedclass · Accepted Answer

$Rs \, 2.16$

Vedclass · Answer

(B) The sides of the triangle are $a = 6 \, cm$,$b = 8 \, cm$,and $c = 10 \, cm$.First,calculate the semi-perimeter $s$:$s = \frac{a + b + c}{2} = \frac{6 + 8 + 10}{2} = \frac{24}{2} = 12 \, cm$.Using Heron's formula,the area of the triangle is:$Area = \sqrt{s(s - a)(s - b)(s - c)}$$Area = \sqrt{12(12 - 6)(12 - 8)(12 - 10)}$$Area = \sqrt{12 	imes 6 	imes 4 	imes 2}$$Area = \sqrt{576} = 24 \, cm^2$.The cost of painting is $9$ paise per $cm^2$,which is $Rs \, 0.09$ per $cm^2$.Total cost $= 24 	imes 0.09 = Rs \, 2.16$.

The edges of a triangular board are $6 \, cm$,$8 \, cm$,and $10 \, cm$. The cost of painting it at the rate of $9$ paise per $cm^2$ is:

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