解：①当点

E

在边AB上时，

∵ \frac{B E}{A B} = \frac{2}{3}, A B = x

，

$∴ A E = \frac{1}{3} x, B E = \frac{2}{3} x,$

$在 R t △ A D E 中, ∴ D E = \frac{\sqrt{10}}{3} x, D H = \frac{\sqrt{10}}{6} x,$

$∵ ∠ A D E = ∠ D G H, ∴ s i n ∠ D G H = \frac{\frac{\sqrt{10}}{6} x}{D G} = \frac{\frac{1}{3} x}{\frac{\sqrt{10}}{3} x},$

$∵ D G = \frac{5}{3} x, ∴ C G = \frac{2}{3} x, ∴ C M = \frac{2}{3} x \cdot t a n ∠ C G M = \frac{2}{9} x,$

$∴ C G = B E,$

在正方形ABCD中： $D C / / A B, ∴ ∠ B = ∠ G C K$ ，

$∵ ∠ C K G = ∠ B K E . ∴ △ C K G ≅ △ B K E . ∴ C K = B K = \frac{1}{2} x$ ，

$∴ M K = \frac{1}{2} x - \frac{2}{9} x = \frac{5}{18} x . ∴ y = \frac{1}{2} \times \frac{5}{18} x \times \frac{2}{3} x = \frac{5}{54} x^{2} .$

②当 $E$ 在AB延长线上时，

$∵ \frac{B E}{A B} = \frac{2}{3}, A B = x, ∴ B E = \frac{2}{3} x, A E = \frac{5}{3} x$ ，

在Rt $△ A D E$ 中， $∵ D E = \frac{\sqrt{34}}{3} x, D H = \frac{\sqrt{34}}{6} x$ ，

$∵ ∠ A D E = ∠ D G H, ∴ s i n ∠ D G H = \frac{\frac{5}{3} x}{\frac{\sqrt{34}}{3} x} = \frac{\frac{\sqrt{34}}{6} x}{D G}$ ，

$∵ D G = \frac{17}{15} x, ∴ C G = \frac{2}{15} x,$

$∴ C M = C G \cdot t a n ∠ C G M = \frac{2}{9} x,$

$∵ ∠ K = ∠ K, ∠ K C G = ∠ K B E = 9 0^{°}, ∴ △ K C G ~ △ K B E,$

$∴ \frac{B E}{C G} = \frac{K B}{K C}, ∴ \frac{\frac{2}{15} x}{C K} = \frac{\frac{2}{3} x}{C K + x}, ∴ C K = \frac{1}{4} x,$

$∴ K M = \frac{1}{4} x + \frac{2}{9} x = \frac{17}{36} x .$

$∴ y = \frac{1}{2} \times \frac{17}{36} x \times \frac{2}{15} x = \frac{17}{540} x^{2} .$

综上所述，当 $\frac{B E}{A B} = \frac{2}{3}$ 时， $∴ y = \frac{5}{54} x^{2}$ 或 $y = \frac{17}{540} x^{2}$ ．

广东省佛山2024年中考一模数学试卷