In the case where you have non-right-angled triangles, different methods are required to figure out the area and angles of the triangle.

## Right-angled triangles In the simple case of a right-angled triangle, the area is given by: $area = \frac{1}{2}bh$

## Using the sin rule

If given the lengths of two sides of a triangle and also the angle joining the two sides the sine rule can be applied to determine the area of a triangle. For a triangle ABC with angles a,b & c shown below: If lengths and b are known and the angle A is known then the area of the triangle is: $area = \frac{1}{2}bc \sin{A}$

## Heron’s formula Given three lengths of a triangle, the area of that triangle can be determined. In Heron’s formula a special variable s is introduced. is the semi-perimeter and is equal to: $s = \dfrac{a + b + c}{2}$

The area of the triangle is therefore: $area = \sqrt{s(s - a)(s - b)(s - c)}$

### Application

Find the area of the following triangle: $s = \dfrac{a + b + c}{2}$ $s = \dfrac{6+5+8}{2}$ $s = 9.5$

Then, use Heron’s formula $area = \sqrt{s(s - a)(s - b)(s - c)}$ $area = \sqrt{9.5(9.5-6)(9.5-5)(9.5-8)}$ $area = 15 units^2$