Question

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

Answer 1

Answer: FOR PARK

a= 120m b= 80m c=50m

s = a+b+c/2

= 120+80+50/2

= 125m

BY HERON'S FORMULA AREA OF PARK

= $\sqrt{s*(s-a)*(s-b)*(s-c)}$

= $\sqrt{125*(125-50)(125-80)(125-120)} \\\sqrt{125*75*45*5}$

= 375$\sqrt{15}$   sq.m

Perimeter of park = 50 + 80 + 120 = 250 m

Thus, length of wire needed = 250 - 3 = 247 m

So, cost of fencing = Rs 20 * 247 = Rs 4940

Step-by-step explanation:

Answer 2

Given : ↴

$\sf \: Sides \: of \: Triangular \: park \: : \: 120 \: m, \: 80 \: m, \: 50 \: m.$

$\sf \: Rate \: of \: barbed \: wire \: at \: the \: rate \: per \: meter \: : Rs. \: 20$

To find : ↴

$\sf \: \rightarrow Area \: of \: the \: Triangular \: park.$

$\sf \: \rightarrow Cost \: of \: fencing \: the \: garden.$

So, Let's Start : ↴

$\sf \: First \: of \: all \: We \: will \: find \: the \: Semi \: perimeter.$

$\sf \red{Semi \: perimeter \: = \: \dfrac{Perimeter}{2} }$

$\sf \longrightarrow \: \dfrac{120 + 80 + 50}{2}$

$\sf \longrightarrow \: \dfrac{250}{2} \: = \: 125 \: meter$

$\boxed {{\sf\purple{Semi \:perimeter \: = \: 125 \: meter}}}$

$\sf \: Now, \: Area \: of \: the \: triangular \: park \: :$

$\sf \underline{By \: Using \: Heron's \: formula \: we \: will \: find \: Area.}$

$\boxed{ \sf \green{Area \: of \: Triangle \: = \: \sqrt{S \: (S - a ) \: (S - b ) \: (S - c )} }}$

$\: \sf \rightarrow \: \sqrt{125(125 - 120) \: (125 - 80) \: (125 - 50)}$

$\: \sf \rightarrow \: \sqrt{125 \: \times \: 5 \: \times \: 45 \: \times \: 75}$

$\: \sf \rightarrow \: \sqrt{25 \: \times \: 5 \: \times 5 \: \: \times \: 9 \: \times \: 25 \: \times \: 3}$

$\sf \rightarrow \: 25 \: \times \: 5 \times \: 3 \: \sqrt{5 \: \times \: 3}$

$\sf \rightarrow \: 375 \: \sqrt{15 } \: m ^{2}$

$\sf \: \blue{So, \: Area \: of \: the \: Triangular \: Plot \: = \: 375 \: \sqrt{15 } \: m ^{2} }$

$\sf \: Also, \: perimeter \: of \: the \: park \: = \: AB \: + \: BC \: + \: CA \: = \: 250 \: meter.$

$\sf \: Therefore, \: Length \: of \: the \: wire \: needed \: for \: fencing :$

$\sf \: 250 \: m \: - \: 3 \: m \: \: = \: 247 \: m$

$\boxed{ \red{\sf \: So, \: the \: Cost \: of \: fencing \: = \: Rs. \: 20 \: \times \: 247 \: = \: Rs. 4940} }$