2023 AMC 12B Problem 11

Let the bases be $a$ and $2a.$ Each leg has horizontal offset $\tfrac{a}{2},$ so the height is $\sqrt{1-\tfrac{a^2}{4}}$ and the area is $A=\tfrac{3a}{2}\sqrt{1-\tfrac{a^2}{4}}.$ Then $A^2=\tfrac94\left(a^2-\tfrac{a^4}{4}\right),$ maximized when $a^2=2.$ There the height is $\tfrac{1}{\sqrt2}$ and $A=\tfrac{3\sqrt2}{2}\cdot\tfrac{1}{\sqrt2}=\tfrac{3}{2}.$ Thus, the correct answer is D.