Area of Isosceles Triangle/Proof 1

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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

IsoscelesTriangleArea.png
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \mathcal A\) \(=\) \(\displaystyle \) \(\displaystyle \frac 1 2 b h\) \(\displaystyle \) \(\displaystyle \)          Area of Triangle in Terms of Side and Altitude          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle \frac 1 2 b \left({r \cos \dfrac \theta 2}\right)\) \(\displaystyle \) \(\displaystyle \)          by definition of cosine          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle \frac 1 2 2 \left({r \sin \dfrac \theta 2}\right) \left({r \cos \dfrac \theta 2}\right)\) \(\displaystyle \) \(\displaystyle \)          by definition of sine          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle \frac 1 2 r^2 \sin \theta\) \(\displaystyle \) \(\displaystyle \)          Double Angle Formula for Sine          

$\blacksquare$