B2. Given is a right-angled triangle $A B C$ with $A B=8, A C=15$ and $B C=17$.

On the side $A B$ lies point $D$ and on the side $A C$ lies point $E$ such that

- the area of $A D E$ is equal to $\frac{3}{4}$ of the area of $A B C$ and
- the perimeter of $A D E$ is equal to the perimeter of $B C E D$.

Calculate the length of $D E$.
B2. Let $A D x$ and $A E y$ then $x y=\frac{3}{4} \times 8 \times 15=90$. For the perimeters, we have: $x+y+D E=(8-x)+(15-y)+D E+17$ so $x+y=20$. Thus, $y=20-x$ and $x^{2}-20 x+90=0$. This gives $x=10-\sqrt{10}$ and $y=10+\sqrt{10} \cdot(10+\sqrt{10})^{2}=110+20 \sqrt{10}$ and $(10-\sqrt{10})^{2}=110-20 \sqrt{10}$ and thus $D E=\sqrt{220}=2 \sqrt{55}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_f71f49936c57e5c73f63g-2.jpg?height=522&width=326&top_left_y=727&top_left_x=1599)
(alternatively: $D E^{2}=x^{2}+y^{2}=(x+y)^{2}-2 x y=20^{2}-2 \times 90=220$ )