Just No Ordinary Triangle - 2

Let the length of one of the legs be $x$ , then the probability of uniformly distributed random variable $X$ is given by $\displaystyle P\left(a \le X \le b \right) = \int_a^b \dfrac {dx}k$ .

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

$\begin{aligned} E\left(A_\triangle \right) & = \int_0^k \frac {x\sqrt{k^2-x^2}}{2k} dx \\ & = \frac k2 \int_0^k \frac xk \sqrt{1-\left(\frac xk\right)^2 } \ dx & \small \color{#3D99F6} \text{Let }\frac xk = \sin \theta, \ dx = k \cos \theta \ d\theta \\ & = \frac {k^2}2 \int_0^\frac \pi 2 \sin \theta \cos^2 \theta \ d\theta \\ & = - \frac {k^2}2 \int_1^0 \cos^2 \theta \ d \cos \theta \\ & = - \frac {k^2}6 \cos^3 \theta \ \bigg|_1^0 \\ & = \frac {k^2}6 \end{aligned}$

For $k=5$ , $E\left(A_\triangle \right) = A = \dfrac {25}6 \approx 4.16667$ , $\implies \lfloor 1000A \rfloor = \boxed{4166}$ .

If the hypotenuse of such triangle is $5$ , then its legs are $x$ and $\sqrt{25-x^2}$ for some length $x$ such that $0 < x < 5$ .

The area of this triangle is $\frac{1}{2} x \sqrt{25 - x^2}$ .

Now, to find its expected area over the inteval $(0, 5)$ , we do

$A = \frac { \displaystyle \int_0^{5} \frac{1}{2} x \sqrt{25 - x^2} \mathrm{d}x }{ 5 - 0 }$

$A = - \frac{1}{10} \cdot \frac{1}{3} (25 - x^2)^{\frac{3}{2}} \Bigg|_0^{5}$

$A = \frac{25}{6} \approx 4.16666667$

Hence, $\lfloor 1000A \rfloor = 4166$ .