# Area of Isosceles Triangle/Proof 1

From ProofWiki

## Theorem

Let $\triangle ABC$ be an isosceles triangle whose apex is $A$.

Let $\theta$ be the angle of the apex $A$.

Let $r$ be the length of a leg of $\triangle ABC$.

Then the area $\mathcal A$ of $\triangle ABC$ is given by:

- $\mathcal A = \dfrac 1 2 r^2 \sin \theta$

## Proof

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \mathcal A\) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \frac 1 2 b h\) | \(\displaystyle \) | \(\displaystyle \) | Area of Triangle in Terms of Side and Altitude | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \frac 1 2 b \left({r \cos \dfrac \theta 2}\right)\) | \(\displaystyle \) | \(\displaystyle \) | by definition of cosine | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \frac 1 2 2 \left({r \sin \dfrac \theta 2}\right) \left({r \cos \dfrac \theta 2}\right)\) | \(\displaystyle \) | \(\displaystyle \) | by definition of sine | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \frac 1 2 r^2 \sin \theta\) | \(\displaystyle \) | \(\displaystyle \) | Double Angle Formula for Sine |

$\blacksquare$