# What is the area of a triangle with sides of length 2, 6, and 5?

Mar 14, 2016

Apply Heron's formula to find the area to be

$\frac{\sqrt{351}}{4} \approx 4.6837$

#### Explanation:

Heron's formula states that, given a triangle with side lengths $a , b , c$ and semiperimeter $s = \frac{a + b + c}{2}$ the area $A$ of the triangle is

$A = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$

In this case, we have $a = 2$, $b = 6$, and $c = 5$. Then, for this triangle we have $s = \frac{2 + 6 + 5}{2} = \frac{13}{2}$. Applying Heron's formula gives us

$A = \sqrt{\frac{13}{2} \left(\frac{13}{2} - 2\right) \left(\frac{13}{2} - 6\right) \left(\frac{13}{2} - 5\right)}$

$= \sqrt{\frac{13}{2} \cdot \frac{9}{2} \cdot \frac{1}{2} \cdot \frac{3}{2}}$

$= \sqrt{\frac{351}{16}}$

$= \frac{\sqrt{351}}{4} \approx 4.6837$