# How do you use Heron's formula to find the area of a triangle with sides of lengths 1 , 1 , and 2 ?

Jan 25, 2016

Heron's formula for finding area of the triangle is given by
$A r e a = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$

Where $s$ is the semi perimeter and is defined as
$s = \frac{a + b + c}{2}$

and $a , b , c$ are the lengths of the three sides of the triangle.

Here let $a = 1 , b = 1$ and $c = 2$

$\implies s = \frac{1 + 1 + 2}{2} = \frac{4}{2} = 2$

$\implies s = 2$

$\implies s - a = 2 - 1 = 1 , s - b = 2 - 1 = 1 \mathmr{and} s - c = 2 - 2 = 0$
$\implies s - a = 1 , s - b = 1 \mathmr{and} s - c = 0$

$\implies A r e a = \sqrt{2 \cdot 1 \cdot 1 \cdot 0} = \sqrt{0} = 0$ square units

$\implies A r e a = 0$ square units

Why is are 0?

The area is 0,because there exists no triangle with the given measurements the given measurements represent a line and a line has no area.

In any triangle the sum of any two sides must be greater than the third side.

If $a , b \mathmr{and} c$ are three sides then
$a + b > c$
$b + c > a$
$c + a > b$

Here $a = 1 , b = 1$ and $c = 2$

$\implies b + c = 1 + 2 = 3 > a$ (Verified)
$\implies c + a = 2 + 1 = 3 > b$ (Verified)
$\implies a + b = 1 + 1 = 2 \cancel{>} c$ (Not Verified)

Since, the property of triangle is not verified therefore, there exists no such triangle.