Finding CDF of the distance from point $T$ to closest diagonal of rectangle A rectangle of dimensions $24\times 10$ is given. A point $T$ is chosen randomly. Find CDF of random variable $X$ that represents distance from point $T$ to closest diagonal of rectangle.
$5$ is the maximum distance from any point to diagonal.
$F_X(t)=  0, t\leq0 $
$F_X(t)=  ?, 0<t\leq5 $ I am not sure how to solve this part...
$F_X(t)=  1, t>5 $
 A: Partition the rectangle into $8$ congruent right triangles, as shown below.

A randomly chosen point in the rectangle is equally likely to be in any of the $8$ triangles, and by symmetry, the distribution of $X$ is the same in each triangle, so we can assume the value of $X$ is based on a randomly chosen point from the lower left corner right triangle, with larger image shown below.

Since for the big right triangle, the area $K$ is given by $K=\frac{1}{2}{\,\cdot\,}12{\,\cdot\,}5=30$, and the length of the hypotenuse is $13$, the length $h$ of the altitude to the hypotenuse is given by $h={\large{\frac{60}{13}}}\approx 4.6$.

For any point in the big right triangle, the distance $d$ to the hypotenuse is in the interval $[0,h]$.

Fixing $d\in [0,h]$, the trapezoidal region above the smaller right triangle is the set of points in the big right triangle whose distance to the hypotenuse is at most $d$.

Let $k$ be the area of the smaller right triangle.$\;$Then by similarity, we get
$$
\frac{k}{K}=\Bigl(\frac{h-d}{h}\Bigl)^2
$$
hence
$$
P(X\le d)=\frac{K-k}{K}=1-\frac{k}{K}=1-\Bigl(\frac{h-d}{h}\Bigr)^2
$$
so
$$
F_X(d)
=
\begin{cases}
0&\text{if}\;\,d < 0\\[4pt]
1-\Bigl({\Large{\frac{h-d}{h}}}\Bigr)^2&\text{if}\;\,0\le d\le h\\[4pt]
1&\text{if}\;\,d > h\\
\end{cases}
$$
where $h={\large{\frac{60}{13}}}$.
