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 $\AA$ of $\triangle ABC$ is given by:
- $\AA = \dfrac 1 2 r^2 \sin \theta$
Proof
\(\ds \AA\) | \(=\) | \(\ds \frac 1 2 b h\) | Area of Triangle in Terms of Side and Altitude | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 b \paren {r \cos \dfrac \theta 2}\) | Definition of Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 2 \paren {r \sin \dfrac \theta 2} \paren {r \cos \dfrac \theta 2}\) | Definition of Sine | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 r^2 \sin \theta\) | Double Angle Formula for Sine |
$\blacksquare$