Area of Isosceles Triangle in terms of Sides

Theorem

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:

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

Proof

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

$\blacksquare$