Question

There is a new colony opened recently in your city and in that colony triangular park has sides 20 m ,

34 m , 42 m.

The gardener has to put a fence all around it and also plant grass inside it.

(i) Find Semi Perimeter of the park.

(ii) Find How much Area does he need to plant the grass.

(iii) Find length of wire needed to fencing the park , If 4 m wide space is to be leave for the gate on

one side.

(iv) Find Cost of fencing it with barbed wire @ ₹ 20 per meter, leaving space 4 m for the gate.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

The three sides of a triangular park are 20m, 34m and 42m

(i) Let we assume that, Three sides be represented as

$\red{\rm :\longmapsto\:a = 20 \: m}$

$\red{\rm :\longmapsto\:b = 34 \: m}$

$\red{\rm :\longmapsto\:c = 42 \: m}$

We know,

$\underline{\boxed{\sf Perimeter \ of \ a \ triangle, \: 2s \: = \: a+b+c}}$

and

$\underline{\boxed{\sf Semi \ perimeter, \: s \: = \: \dfrac{Perimeter}{2} \: }}$

$\rm :\longmapsto\:s \: = \: \dfrac{a + b + c}{2}$

$\rm :\longmapsto\:s \: = \: \dfrac{20 + 34+ 42}{2}$

$\rm :\longmapsto\:s \: = \: \dfrac{96}{2}$

$\bf\implies \:s = 48 \: m$

(ii) Now, Required area to plant the grass is evaluated using Heron's Formula, which is given by

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

Now, on Substituting the value of s, a, b and c, we get

$\rm :\longmapsto\:Area_{triangular \: park} \: = \sqrt{48(48 - 20)(48 - 24)(48 - 42)}$

$\rm :\longmapsto\:Area_{triangular \: park} \: = \sqrt{48 \times 28 \times 14 \times 6}$

$\rm :\longmapsto\:Area_{triangular \: park} \: = \sqrt{16 \times 3 \times 7 \times 4 \times 7 \times 2 \times 3 \times 2}$

$\rm :\longmapsto\:Area_{triangular \: park} \: = 4 \times 4 \times 7 \times 3$

$\rm :\longmapsto\:Area_{triangular \: park} \: = 336 \: {m}^{2}$

(iii) Now, Required length of wire needed to fencing the park, leaving 4 m space for gate is

$\rm \: = \: \: a + b + c - 4$

$\rm \: = \: \: 20 + 34 + 42 - 4$

$\rm \: = \: \: 96 - 4$

$\rm \: = \: \: 92 \: m$

(iv) Cost of fencing it with barbed wire @ ₹ 20 per meter, leaving 4 m space for gate.

Now, it is given that

Cost of fencing 1 meter is ₹ 20

So,

Cost of fencing 92 meter is 20 × 92 = ₹ 1840