Diagonal of a rectangle.

Let $x$ and $y$ denote the sides of the rectangle.

We have,

$\begin{aligned} 0&<x\leq a \hspace{4mm}\color{#3D99F6}\small (1)\\ 0&<y\leq a \hspace{4mm}\color{#3D99F6}\small (2)\\ \text{Let length of the diagonal be d}\\ d&=\sqrt{x^2+y^2}\\ \text{According to the problem we want,}\\ d&<a\\ \implies\sqrt{x^2+y^2}&<a\\ x^2+y^2&<a^2 \hspace{4mm}\color{#3D99F6}\small (3)\\ \text{From} \color{#3D99F6}(1) ,(2) \color{#333333}\text{ and } \color{#3D99F6} (3)\end{aligned}$

We can see that the desired region reduces to a quarter circle of radius $a$ ,

while the sample space is a square of side $a$

Thus the required probablity is given by,

$\begin{aligned} p&=\dfrac{\text{Area of quarter circle}}{\text{Area of square }}=\dfrac{\dfrac{\pi a^2}{4}}{a^2}\\ &=\dfrac{\pi}{4}\\ &\approx \color{#EC7300}\boxed{\color{#333333}0.7853}\end{aligned}$