
Area:
$A = xy$
By similar triangle:
$\cot \alpha = \dfrac{r}{x} = \dfrac{b}{a}$
$r = \dfrac{b}{a}x$
$\cot \beta = \dfrac{s}{x} = \dfrac{a}{b}$
$s = \dfrac{a}{b}x$
$r + y + s = \sqrt{a^2 + b^2}$
$\frac{b}{a}x + y + \dfrac{a}{b}x = \sqrt{a^2 + b^2}$
$y + \dfrac{a^2 + b^2}{ab}x = \sqrt{a^2 + b^2}$
$y = \sqrt{a^2 + b^2} - \dfrac{a^2 + b^2}{ab}x$
$y = \dfrac{ab\sqrt{a^2 + b^2} - (a^2 + b^2)x}{ab}$
$y = \dfrac{\sqrt{a^2 + b^2}}{ab} \left( ab - x \sqrt{a^2 + b^2} \right)$
Thus,
$A = x \dfrac{\sqrt{a^2 + b^2}}{ab} \left( ab - x \sqrt{a^2 + b^2} \right)$
$A = \dfrac{\sqrt{a^2 + b^2}}{ab} \left( abx - x^2 \sqrt{a^2 + b^2} \right)$
$\dfrac{dA}{dx} = \dfrac{\sqrt{a^2 + b^2}}{ab} \left( ab - 2x \sqrt{a^2 + b^2} \right) = 0$
$2x \sqrt{a^2 + b^2} = ab$
$x = \dfrac{ab}{2 \sqrt{a^2 + b^2}}$
$y = \dfrac{\sqrt{a^2 + b^2}}{ab} \left( ab - \dfrac{ab}{2 \sqrt{a^2 + b^2}} \cdot \sqrt{a^2 + b^2} \right)$
$y = \dfrac{\sqrt{a^2 + b^2}}{ab} \left( ab - \dfrac{ab}{2} \right)$
$y = \dfrac{\sqrt{a^2 + b^2}}{ab} \left( \dfrac{ab}{2} \right)$
$y = \dfrac{\sqrt{a^2 + b^2}}{2}$
Dimensions:
$\dfrac{ab}{2 \sqrt{a^2 + b^2}} \, \times \, \dfrac{\sqrt{a^2 + b^2}}{2}$ answer