Let $t$ be TNYWR.

Trapezoid $A B C D$ has $\angle A D C=\angle B C D=90^{\circ}, A D=t, B C=4$, and $C D=t+13$. What is the length of $A B$ ?

![](https://cdn.mathpix.com/cropped/2024_04_17_6105d4c4bb695ce107edg-3.jpg?height=404&width=477&top_left_y=2107&top_left_x=821)
Let $E$ be on $A D$ so that $B E$ is perpendicular to $A E$.

![](https://cdn.mathpix.com/cropped/2024_04_17_54c374478de6843636d1g-24.jpg?height=406&width=544&top_left_y=171&top_left_x=888)

Quadrilateral $B C D E$ has three right angles, which means that it must have four right angles.

This means that quadrilateral $B C D E$ is a rectangle, so $B E=C D=t+13$.

We also have that $E D=B C=4$, so $A E=A D-D E=t-4$.

By the Pythagorean theorem applied to $\triangle A E B$, we get

$$
A B^{2}=(t-4)^{2}+(t+13)^{2}=2 t^{2}+18 t+185
$$

Since $A B$ is positive, $A B=\sqrt{2 t^{2}+18 t+185}$.

Since $t=11$,

$$
A B=\sqrt{2(121)+18(11)+185}=\sqrt{242+198+185}=\sqrt{625}=25
$$