A Triangular Park ABC has sides 120 m, 80m and 50m. a gardener has to put a fence all around it and also plant grass inside. How much area does she need to plant? Find the cost of fencing it with barbed wire at the rate of of Rs. 20 per metre leaving a space 3m wide for a gate on one side?​

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$\sf\underline\bold{Step-by-Step:}$

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$\sf\bold{Given(In \:the\:Q):}$

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$\sf\bold{To \: find:}$

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$\sf\small{☆Area\:to\:be\:planted=Area \: of \: ∆ABC}$

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$\sf\underline\bold{Calculating\:area\:of\:∆ABC:}$

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Use heron's formula to find the area of the triangle.

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$\mapsto$ $\sf\small{Heron's\:formula=}$

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$\sf\bold{Now,find\:semi\:parameter:-}$

$\sf\small{Perimeter\:of\:the\:∆=120+80+50=250}$

$\sf\small{Semi-Perimeter:}$ $\sf\dfrac{250}{2}$ $\sf\small{=125m}$

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$\sf\small{Substitute \: the\:values\:in\:heron's\:formula:}$

$\sf{Area\:of\:the\:∆:-}$

$\mapsto$

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$\mapsto$ $\sf\sqrt{125\times(5)\times(45)\times(75)}$

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$\mapsto$ $\sf\small\sqrt{2109375}$ $\sf\small{=375}$ $\sf\small\sqrt{15}$

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$\longmapsto$ $\sf\underline\bold\purple{1452.56m^2}$

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$\sf\underline\bold{Now,find\:the\:cost\:of\:fencing:}$

$\sf{Cost\:of\:fencing-}$

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$\sf\underline{Hence,the\:gardener\:has\:to\:fence:}$

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So,total cost of fencing at the rate of Rs.20 per m:-

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f(x) has value 150 at x= 0

and 0 at x= 3

so d(f(x))/dt = slope = -150/3 = -50

the answer is 62.8 cubic units