Maximum Area of Isosceles Triangle

Theorem

Consider two line segments $A$ and $B$ of equal length $a$ which are required to be the legs of an isosceles triangle $T$.

Then the area of $T$ is greatest when the apex of $T$ is a right angle.

The area of $T$ in this situation is equal to $\dfrac {a^2} 2$.

Let $\triangle OAB$ be the isosceles triangle $T$ formed by the legs $OA$ and $OB$.

Thus the apex of $T$ is at $O$.

Let $\theta$ be the angle $\angle AOB$.

We see that by keeping $OA$ fixed, $B$ can range over the semicircle $AOB$.

Thus $\theta$ can range from $0$ to $180 \degrees$, that is, $2$ right angles.

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

This is a maximum when $\sin \theta = 1$, that is, when $\theta$ is a right angle.

The result follows.

$\blacksquare$

1992: David Wells: Curious and Interesting Puzzles ... (previous) ... (next): The Area Enclosed Against The Seashore: $33$