Let $t$ be TNYWR.

In the diagram, $\angle B A E=\angle C B E=\angle D C E=90^{\circ}$. If $A E=\sqrt{5}, A B=\sqrt{4}, B C=\sqrt{3}$, and $C D=\sqrt{t}$, what is the length of $D E$ ?

![](https://cdn.mathpix.com/cropped/2024_04_30_4efeb798298dfec26b46g-2.jpg?height=431&width=360&top_left_y=1465&top_left_x=907)
We use the Pythagorean Theorem repeatedly and obtain

$$
D E^{2}=C D^{2}+C E^{2}=C D^{2}+\left(B C^{2}+B E^{2}\right)=C D^{2}+B C^{2}+\left(A B^{2}+A E^{2}\right)
$$

Since $A E=\sqrt{5}, A B=\sqrt{4}, B C=\sqrt{3}$, and $C D=\sqrt{t}$, then

$$
D E^{2}=t+3+4+5=t+12
$$

Since the answer to (b) is 4 , then $t=4$, and so $D E^{2}=16$.

Since $D E>0$, then $D E=\sqrt{16}=4$.

ANSWER: $6,4,4$