# If sides A and B of a triangle have lengths of 2 and 4 respectively, and the angle between them is (pi)/4, then what is the area of the triangle?

Dec 13, 2015

$2 \sqrt{2}$

#### Explanation:

One way to do this is strictly geometrical.

Consider the square containing the given triangle with the side of length $4$ as the diagonal.
We would have the structure below:

We are asked for the area of $\triangle T$
and it is clearly
$\textcolor{w h i t e}{\text{XXX}}$the area of the $\square$
$\textcolor{w h i t e}{\text{XXX}}$minus
$\textcolor{w h i t e}{\text{XXX}}$(the area of $\triangle A$ plus $\triangle B$)

With a diagonal of length $4$ each side of the square has a length of $2 \sqrt{2}$ (as shown) based on the Pythagorean Theorem.
Therefore the area of the $\square$ is $2 \sqrt{2} \times 2 \sqrt{2} = 8$

The area of $\triangle A$ is ($\frac{1}{2} b h$)
$\textcolor{w h i t e}{\text{XXX}} \frac{1}{2} \times 2 \sqrt{2} \times 2 = 4$

The area of $\triangle B$ is
$\textcolor{w h i t e}{\text{XXX}} \frac{1}{2} \times \left(2 \sqrt{2} - 2\right) \times 2 \sqrt{2} = 4 - 2 \sqrt{2}$

The area of $\triangle T$ is
$\textcolor{w h i t e}{\text{XXX}} 8 - \left(4 + 4 - 2 \sqrt{2}\right) = 2 \sqrt{2}$