Area of Isosceles Triangle in terms of Sides

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

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

Let $b$ be the length of the base of $\triangle ABC$.

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

$\AA = \dfrac b 4 \sqrt {4 r^2 - b^2}$



Let $h$ be the height of $\triangle ABC$.

\(\ds \AA\) \(=\) \(\ds \frac 1 2 b h\) Area of Triangle in Terms of Side and Altitude
\(\ds \) \(=\) \(\ds \frac b 2 \sqrt {r^2 - \paren {\frac b 2}^2}\) Pythagoras's Theorem
\(\ds \) \(=\) \(\ds \frac b 2 \sqrt {\frac {4 r^2 - b^2} 4}\) simplification
\(\ds \) \(=\) \(\ds \frac b 4 \sqrt {4 r^2 - b^2}\) simplification