Area of Isosceles Triangle

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 1

\(\ds \AA\)

\(=\)

\(\ds \frac 1 2 b h\)

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

$\blacksquare$

Proof 2

A direct application of Area of Triangle in Terms of Two Sides and Angle:

$\AA = \dfrac 1 2 a b \sin \theta$

where $a = b = r$.

$\blacksquare$