Area of a Triangle

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 c 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. s 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:

First, find the value of s

$s = \dfrac{a + b + c}{2}$
$s = \dfrac{6+5+8}{2}$
$s = 9.5$

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

See also:

Sine rule

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