Find the area of the triangle in which $a = 52 c m, b = 56 c m, c = 60 c m$ , where $a, b, c$ are sides of triangle.

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Solution

Given:-
$a = 52 c m$
$b = 56 c m$
$c = 60 c m$
Semi-perimeter of triangle $(s) = \frac{a + b + c}{2}$
$\Rightarrow s = \frac{52 + 56 + 60}{2} c m = 84 c m$
Now, by using Heron's formula-
Area of triangle
$= \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{84 (84 - 52) (84 - 56) (84 - 60)} c m^{2} = \sqrt{84 \times 32 \times 28 \times 24} c m^{2} = 1344 {c m}^{2}$

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