A triangular park ABC has sides 120 m, 80 m and 50 m (as shown in the adjoining figure). A gardner Dhania has to put a fence 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 ₹ 20 per metre leaving a space 3 m wide for a gate on one side.

It is given that,

ABC is a triangular park with sides 120 m, 80 m and 50 m.

Here, the perimeter of △ABC = 120 + 80 + 50 = 250 m

Portion at which a gate is built = 3m

Remaining perimeter = 250 – 3 = 247 m.

So, the length of the fence around it = 247 m.

Rate of fencing = ₹20 per metre

Total cost of fencing = 20 × 247 = ₹4940.

We know that,

Semi perimeter (s) = $\dfrac{\text{Perimeter}}{2} = \dfrac{250}{2}$ = 125 cm.

Area of triangle = $\sqrt{s(s - a)(s - b)(s - c)}$

Substituting values we get,

$A = \sqrt{125(125 - 120)(125 - 80)(125 - 50)} \\[1em] = \sqrt{125 \times 5 \times 45 \times 75} \\[1em] = \sqrt{(25 \times 5) \times 5 \times (5 \times 3 \times 3) \times (25 \times 3)} \\[1em] = \sqrt{(25)^2 \times 5 \times (5)^2 \times (3)^2 \times (3)} \\[1em] = 25 \times 5 \times 3 \times \sqrt{5 \times 3} \\[1em] = 375\sqrt{15} \text{ m}^2.$

Hence, area needed for plantation = $375\sqrt{15}$ m² and cost of fencing = ₹4940.