Question

the sides of the triangular field are 55m, 300m, 300m.its area os equal to 7/15 the area ofvrhe circular park. what is the Perimeter of the circular park​

Answer 1

Given:

The sides of the triangular field are 55m, 300m, 300m.

Its area is equal to 7/15 the area of the circular park.

To find:

What is the perimeter of the circular park​?

Solution:

Finding the area of the triangular field:

We know the Heron's formula of a triangle with sides a, b & c is as follows

$\boxed{\bold{Semi-Perimeter, S = \frac{a+b+c}{2} }}\\\\\boxed{\bold{Area = \sqrt{S(S-a)(S-b)(S-c)} }}$

Here, a = 55 m, b = 300 m & c = 300 m

∴ S = $\frac{55+ 300+300}{2} = \frac{655}{2} = 327.5 \:m$

Then,

$Area$ = $\sqrt{327.5 (327.5 - 55)(327.5 - 300)(327.5 - 300)}$

$\implies Area = \sqrt{327.5 \times 272.5 \times 27.5 \times 27.5}$

$\implies Area = 27.5 \times \sqrt{327.5 \times 272.5}$

$\implies Area = 27.5 \times 298.73$

$\implies \bold{Area = 8215.075\:m^2}$ ← area of the triangular park

Finding the area of the circular park:

It is given that,

Area of the triangular park = $\frac{7}{15} \times$ Area of the circular park

substituting the value of the area of the triangular park

⇒ 8215.075 = $\frac{7}{15} \times$ Area of the circular park

⇒ Area of the circular park = $\frac{15\times 8215.975}{7}$

⇒ Area of the circular park = $\bold{17603.73\:m^2}$

Finding the radius of the circular park:

Let the radius of the circular park be "r".

Area of the circular park = $\bold{17603.73\:m^2}$

$\implies \pi r^2 = 17603.73$

$\implies \frac{22}{7} r^2 = 17603.73$

$\implies r^2 = \frac{17603.73\times 7}{22}$

$\implies r = \sqrt{5601.18}$

$\implies \bold{r = 74.84\:m}$

Finding the perimeter of the circular park:

The perimeter of the circular park is,

= Perimeter of the circle

= $2\pi r$

substituting the value of r = 74.84

= $2 \times \frac{22}{7} \times 74.84$

= $\bold{470.42\:m^2}$

Thus, the perimeter of the circular park is → 470.42 m².

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