Question

find the area of the triangle whose sides are 122 m , 120 m, 22m. (hints - are = ( s ( s-a ) ( s-b ) ( s-c ))^1/2) where a b c are sides of the triangle and s is semi perimeter of the triangle

Answer 1

hi! Astha here is your answer

Answer 2

Hey friend,
Here is the answer you were looking for:

Given that,
The sides of Triangle are : 122m, 120m and 22m

Let, a = 122m
b = 120m
c = 22m

We know that s (semi-perimeter) = a + b + c/2

So,

$s = \frac{122 + 120 + 22}{2} \\ \\ s = \frac{264}{2} \\ \\ s = 132m$

We know that
The Area of triangle (According to Heron's Formula) = √(s(s - a)(s - b)(s - c)

So,
putting the values and calculating :

$\sqrt{132(132 - 122)(132 - 120)(132 - 22)} \\ \\ = \sqrt{132(10)(12)(110)} \\ \\ = \sqrt{1742400} \\ \\ = 1320 {cm}^{2}$


Hope this helps!!

If you have any doubt regarding to my answer, feel free to ask in the comment section or inbox me if needed.

@Mahak24

Thanks...
☺☺