# How do you use Heron's formula to determine the area of a triangle with sides of that are 5, 6, and 3 units in length?

$A r e a = 2 \sqrt{14}$ square units

$A r e a = 7.48331$ square units

#### Explanation:

The Heron's formula to determine the area of a triangle is:

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

where $s = \frac{1}{2} \left(a + b + c\right)$

$s$ is one-half of the perimeter of the triangle.

To compute for the area of the triangle using the Heron's Formula, the $s$ should be computed first.

Since the given sides are $a = 5$, $b = 6$, and $c = 3$.

$s = \frac{1}{2} \cdot \left(5 + 6 + 3\right) = 7$

$s = 7$

Compute the Area after computing s:

$A r e a = \sqrt{7 \left(7 - 5\right) \left(7 - 6\right) \left(7 - 3\right)}$

$A r e a = \sqrt{7 \left(2\right) \left(1\right) \left(4\right)}$

$A r e a = 2 \sqrt{14}$

$A r e a = 7.48331$ square units

Have a nice day!!! from the Philippines ....

Jun 15, 2018

There's always a better alternative than Heron's Formula. Area $S$ satisfies

$16 {S}^{2} = \left(a + b + c\right) \left(- a + b + c\right) \left(a - b + c\right) \left(a + b - c\right) = \left(5 + 6 + 3\right) \left(- 5 + 6 + 3\right) \left(5 - 6 + 3\right) \left(5 + 6 - 3\right) = 14 \left(4\right) \left(2\right) \left(8\right)$ or

$S = 2 \sqrt{14} .$