The perimeter of a triangle is 300 m. If its sides are in the ratio of 3: 5: 7. Find the area of the triangle.

Question:

The perimeter of a triangle is 300 m. If its sides are in the ratio of 3: 5: 7. Find the area of the triangle.

Solution:

Given the perimeter of a triangle is 300 m and the sides are in a ratio of 3: 5: 7

Let the sides a, b, c of a triangle be 3x, 5x, 7x respectively

So, the perimeter = 2s = a + b + c

200 = a + b + c

300 = 3x + 5x + 7x

300 = 15x

Therefore, x = 20 m

So, the respective sides are

a = 60 m

b = 100 m

c = 140 m

Now, semi perimeter

$s=\frac{a+b+c}{2}$

$=\frac{60+100+140}{2}$

= 150 m

By using Heron's Formula

The area of a triangle $=\sqrt{s \times(s-a) \times(s-b) \times(s-c)}$

$=\sqrt{150 \times(150-60) \times(150-100) \times(150-140)}$

$=1500 \sqrt{3} \mathrm{~m}^{2}$

Thus, the area of a triangle is $1500 \sqrt{3} \mathrm{~m}^{2}$

 

Leave a comment