Just No Ordinary Triangle

If the hypotenuse of such triangle is $5$ , then its legs are $5 \sin \theta$ and $5 \cos \theta$ for some angle $\theta$ such that $0 < \theta < \frac{\pi}{2}$ .

The area of this triangle is $\frac{25}{2} \sin \theta \cos \theta$ .

Now, to find its expected area over the inteval $(0, \frac{\pi}{2})$ , we do

$A = \frac { \displaystyle \int_0^{\frac{\pi}{2}} \frac{25}{2} \sin \theta \cos \theta \mathrm{d}\theta }{ \frac{\pi}{2} - 0 }$

$A = \frac{25}{\pi} \frac{\sin^2 \theta}{2} \Bigg|_0^{\frac{\pi}{2}}$

$A = \frac{25}{2\pi} \approx 3.978873$

Hence, $\lfloor 1000A \rfloor = 3978$ .

Let one of the non-right angle be $x$ , then the probability of uniformly distributed random variable $X$ is given by $\displaystyle P\left(a \le X \le b \right) = \int_a^b \dfrac 1{\frac \pi 2} dx = \int_a^b \dfrac 2 \pi \ dx$ .

Since the area of the right triangle with hypotenuse of $k$ is $A_\triangle = \dfrac {k^2\sin x \cos x}2 = \dfrac {k^2\sin 2x}4$ , then the expected value of $A_\triangle$ is as follows:

$\begin{aligned} E\left(A_\triangle \right) & = \int_0^\frac \pi 2 \frac 2\pi \cdot \frac {k^2\sin 2x}4 dx = - \frac {k^2\cos 2x}{4\pi} \ \bigg|_0^\frac \pi 2 = \frac {k^2}{2\pi} \end{aligned}$

For $k=5$ , $E\left(A_\triangle \right) = A = \dfrac {25}{2\pi} \approx 3.97887$ , $\implies \lfloor 1000A \rfloor = \boxed{3978}$ .