2019 AMC 10B Problem 15

Let the two shared side lengths be $a\le b$. Because the triangles have different areas, the shared sides cannot play the same roles in both triangles. Thus the area-$2$ triangle has legs $a$ and $b$, while the area-$1$ triangle has leg $a$, other leg $\sqrt{b^2-a^2}$, and hypotenuse $b$. The product of the two non-shared third sides is $\sqrt{a^2+b^2}\sqrt{b^2-a^2}$, whose square is $b^4-a^4$. Using the areas, $\dfrac{ab}{2}=2$ and $\dfrac{a\sqrt{b^2-a^2}}{2}=1$. Hence $a^2b^2=16$ and $a^2(b^2-a^2)=4$, so $a^4=12$. Then $b^4=\dfrac{256}{12}=\dfrac{64}{3}$, and $b^4-a^4=\dfrac{28}{3}$. Thus, A is the correct answer.