# How do you use Heron's formula to find the area of a triangle with sides of lengths 9 , 3 , and 9 ?

Jan 24, 2016

$A = \frac{9 \sqrt{35}}{4} \approx 13.3112$

#### Explanation:

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

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

Here, we know that

$s = \frac{9 + 3 + 9}{2} = \frac{21}{2}$

which gives an area of

$A = \sqrt{\frac{21}{2} \left(\frac{21}{2} - 9\right) \left(\frac{21}{2} - 3\right) \left(\frac{21}{2} - 9\right)}$

$A = \sqrt{\frac{21}{2} \left(\frac{3}{2}\right) \left(\frac{15}{2}\right) \left(\frac{3}{2}\right)}$

$A = \sqrt{\frac{{9}^{2} \times 7 \times 5}{4} ^ 2}$

$A = \frac{9 \sqrt{35}}{4} \approx 13.3112$