Foundations, part 15

There are two cases: the hypotenuse is twice the length of a leg, or one leg is twice the length of the other leg. The first case gives legs of length $\dfrac{H}{2}$ and $\dfrac{H\sqrt{3}}{2}$ , so the area would be $\dfrac{H^2\sqrt{3}}{8}$ . However, that's not one of the answer choices, so we turn to the second case: one leg is twice the length of the other. Here, our leg lengths are $\dfrac{H}{\sqrt{5}}$ and $\dfrac{2H}{\sqrt{5}}$ . So, the area is $\boxed{\dfrac{H^2}{5}}$ .

(I'm typing on mobile right now, so I'll expand on this solution later.)

Let the two legs of the right-angled triangle have lengths $a$ and $2a$ . By the Pythagoras's Theorem,

$a^2 + (2a)^2 = H^2$

$5a^{2} = H^2$ .

Therefore,

Area $= \frac{1}{2} \times a \times 2a = a^2 = \boxed{\frac{H^2}{5}}$