SAT Question f811d345: Right Triangles And Trigonometry answer and explanation with Desmos solution

It's given that the legs of a right triangle have lengths $24$ centimeters and $21$ centimeters. In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the two legs. It follows that if $h$ represents the length, in centimeters, of the hypotenuse of the right triangle, $h^2=24^2+21^2$. This equation is equivalent to $h^2=1,017$. Taking the square root of each side of this equation yields $h=\sqrt{1,017}$. This equation can be rewritten as $h=\sqrt{9\cdot 113}$, or $h=\sqrt{9}\cdot \sqrt{113}$. This equation is equivalent to $h=3\sqrt{113}$. It's given that the length of the triangle's hypotenuse, in centimeters, can be written in the form $3\sqrt{d}$. It follows that the value of $d$ is $113$.