A triangular park has sides $120 m, 80 m a n d 50 m$ . 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 $R s 30 p e r m$ leaving a space of $3 m$ wide for a gate on one side.

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Solution

Step 1: Finding the area of the triangular park, using the Heron's formula
$\begin{array}{l} A = \sqrt{s (s - a) (s - b) (s - c)} \\ a = 50 m, b = 80 m, c = 120 m \\ s = \frac{a + b + c}{2} \\ s = \frac{50 + 80 + 120}{2} \\ s = \frac{250}{2} \\ s = 125 m \\ A = \sqrt{s (s - a) (s - b) (s - c)} \\ A = \sqrt{125 (125 - 50) (125 - 80) (125 - 120)} \\ A = \sqrt{125 (75) (45) (5)} \\ A = 375 \sqrt{15} m^{2} \end{array}$
Step 2: Find the perimeter of the park
$\begin{array}{l} P e r i m e t e r = a + b + c \\ P e r i m e t e r = 50 + 80 + 120 \\ P e r i m e t e r = 250 m \end{array}$
Step 3: Find the cost of fencing
Thus, length of wire needed $= 250 - 3 = 247$ m
So, cost of fencing $= 30 \times 247 = R s 7410$
The area needs to plant is $375 \sqrt{15} m^{2}$ and the cost of fencing is $R s 7410$

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