In the kite $ABCD$, the two shorter sides are: $AB=AD=a$, and the other two sides are: $CB=CD=b$; the shorter diagonal is $BD=d$. Parallel lines are drawn at equal distances from the diagonal $BD$ on both sides. These two parallel lines define the inscribed quadrilateral $EFGH$ in the kite. How does the area of this quadrilateral change if the two parallel lines move away from $BD$ in equal measure? Characterize and graph this change! (Given values: $a=5, b=3\sqrt{5}, d=6$ units.)
Let $E F\|H G\| B D$ and $O A=m_{1}, O C=m_{2}$. The $E F G H$ is an isosceles trapezoid (symmetric with respect to the $A C$ axis). If for brevity $E F=u, H G=v$, then considering that the height of the trapezoid is $2 x$, the area of the trapezoid is

$$
y=\frac{u+v}{2} \cdot 2 x=u x+v x
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_cd4c983b87df443a2402g-1.jpg?height=474&width=374&top_left_y=370&top_left_x=865)

$$
A B D \Delta \sim A E F \Delta ; \quad \text { therefore } \quad B D: E F=O A: I_{1} A
$$

thus

$$
\begin{aligned}
d: u & =m_{1}:\left(m_{1}-x\right), \quad \text { so } \quad u=d-\frac{d}{m_{1}} x \\
B C D \Delta & \sim H G C \Delta ; \quad \text { therefore } \quad B D: H G=O C: I_{2} A
\end{aligned}
$$

thus

$$
d: v=m_{2}:\left(m_{2}-x\right)
$$

and hence

$$
v=d-\frac{d}{m_{2}} x
$$

Based on these,

$$
y=2 d x-\frac{d\left(m_{1}+m_{2}\right)}{m_{1} m_{2}} x^{2}
$$

We need to examine this function for $0 \leq x \leq m_{1}$. (Indeed, $m_{1} < m_{2}$)

| $\max$ | 14 | 13.5 | 11.1 | 8 |

![](https://cdn.mathpix.com/cropped/2024_05_02_cd4c983b87df443a2402g-2.jpg?height=515&width=352&top_left_y=127&top_left_x=884)