25.(10分)(2024·句容期末)如图①,四边形ABCD中,$AD// BC,∠B= 90^{\circ },BC= 2AD= 4$.点P从点C出发,沿着折线CBA运动,到达点A停止运动.设点P运动的速度为每秒1个单位长度,运动时间为x s,连接DP,记$\triangle DPC$的面积为y.已知点P在线段CB上运动时满足的函数关系式是$y= \frac {3}{2}x$.请解答下列问题:
(1)$AB= $______;
(2)当点P在BA上运动时,求y关于x的函数表达式,并在图②中补全图象;
(3)结合图象,当$\triangle DPC$的面积不小于四边形ABCD面积的$\frac {4}{9}$时,求出x的取值范围.
(1)$AB= $______;
(2)当点P在BA上运动时,求y关于x的函数表达式,并在图②中补全图象;
(3)结合图象,当$\triangle DPC$的面积不小于四边形ABCD面积的$\frac {4}{9}$时,求出x的取值范围.
答案：
(1) 3(2) 解：如答图①，点$P$在$AB$上运动，$\because AD // BC$，$\angle B = 90^{\circ}$，
$\therefore \angle A = 180^{\circ} - \angle B = 90^{\circ}$，$\therefore PB \perp BC$，$PA \perp AD$，
$\because BC = 2AD = 4$，$\therefore AD = 2$。
由(1)知$AB = 3$，$\therefore AB + BC = 3 + 4 = 7$，
$\therefore BP = x - 4$，$AP = 7 - x$。
$\because S_{\triangle DPC} = S_{四边形ABCD} - S_{\triangle PBC} - S_{\triangle PAD} = \frac{1}{2}(AD + BC)·AB - \frac{1}{2}BC·BP - \frac{1}{2}AD·AP$，
$\therefore y = \frac{1}{2}×(2 + 4)×3 - \frac{1}{2}×4(x - 4) - \frac{1}{2}×2(7 - x) = -x + 10$。
当点$P$与点$B$重合时，$x = 4$；当点$P$与点$A$重合时，$x = 7$，
$\therefore y = -x + 10(4 \leq x \leq 7)$，补全函数图象如答图②。
(3) 解：$\because S_{四边形ABCD} = \frac{1}{2}(AD + BC)·AB = \frac{1}{2}×(2 + 4)×3 = 9$，$\therefore \frac{4}{9}S_{四边形ABCD} = \frac{4}{9}×9 = 4$，
$\because S_{\triangle DPC} \geq \frac{4}{9}S_{四边形ABCD}$，即$y \geq 4$，
$\therefore \begin{cases}\frac{3}{2}x \geq 4\\-x + 10 \geq 4\end{cases}$，解得$\frac{8}{3} \leq x \leq 6$，
$\therefore x$的取值范围是$\frac{8}{3} \leq x \leq 6$。
(1) 3(2) 解：如答图①，点$P$在$AB$上运动，$\because AD // BC$，$\angle B = 90^{\circ}$，
$\therefore \angle A = 180^{\circ} - \angle B = 90^{\circ}$，$\therefore PB \perp BC$，$PA \perp AD$，
$\because BC = 2AD = 4$，$\therefore AD = 2$。
由(1)知$AB = 3$，$\therefore AB + BC = 3 + 4 = 7$，
$\therefore BP = x - 4$，$AP = 7 - x$。
$\because S_{\triangle DPC} = S_{四边形ABCD} - S_{\triangle PBC} - S_{\triangle PAD} = \frac{1}{2}(AD + BC)·AB - \frac{1}{2}BC·BP - \frac{1}{2}AD·AP$，
$\therefore y = \frac{1}{2}×(2 + 4)×3 - \frac{1}{2}×4(x - 4) - \frac{1}{2}×2(7 - x) = -x + 10$。
当点$P$与点$B$重合时，$x = 4$；当点$P$与点$A$重合时，$x = 7$，
$\therefore y = -x + 10(4 \leq x \leq 7)$，补全函数图象如答图②。
(3) 解：$\because S_{四边形ABCD} = \frac{1}{2}(AD + BC)·AB = \frac{1}{2}×(2 + 4)×3 = 9$，$\therefore \frac{4}{9}S_{四边形ABCD} = \frac{4}{9}×9 = 4$，
$\because S_{\triangle DPC} \geq \frac{4}{9}S_{四边形ABCD}$，即$y \geq 4$，
$\therefore \begin{cases}\frac{3}{2}x \geq 4\\-x + 10 \geq 4\end{cases}$，解得$\frac{8}{3} \leq x \leq 6$，
$\therefore x$的取值范围是$\frac{8}{3} \leq x \leq 6$。