$\sf\underline\bold{Answer:}$
- $\sf\small\underline{\underline{Area\: planted\: by\: the\: gardener : 1452.36m^2}}$
- $\sf\small\underline{\underline{The\:cost\:of\:fencing\:the\:park:Rs.4940}}$
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$\sf\underline\bold{Step-by-Step:}$
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$\sf\bold{Given(In \:the\:Q):}$
- Sides of the triangular park are 120m,80m and 50m.
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$\sf\bold{To \: find:}$
- How much area of the park does she need to plant?
- The cost of fencing the park ?
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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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$\mapsto$ $\sf\small{Heron's\:formula=}$
[tex]\sf\sqrt{s(s-a)(s-b)(s-c)}[/tex]
- $\sf{Where\:s=semi\:perimeter}$
- $\sf{a,b,c\: = side\:of\:the\:∆}$
- $\sf\small{Here\:a=120,b=80 \:and\: c=50}$
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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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$\sf\small{Substitute \: the\:values\:in\:heron's\:formula:}$
$\sf{Area\:of\:the\:∆:-}$
$\mapsto$ [tex]\sf\sqrt{125(125-120)(125-80)(125-50)}[/tex]
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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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$\sf\underline\bold{Now,find\:the\:cost\:of\:fencing:}$
$\sf{Cost\:of\:fencing-}$
- $\sf{Rate = Rs.20 per \:meter}$
- $\sf{Left\:space=3m}$
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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:-
- $\sf\underline\bold\purple{=247\times20=4940}$
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