《2026年龙江王中王中考总复习数学》

20. (2025 江苏连云港)一块直角三角形木板,它的直角边$BC$长$2m$,面积为$1.5m^{2}$.
(1)甲、乙两人分别按图①、图②用它设计一个正方形桌面,请说明哪个正方形面积较大;
(2)丙、丁两人分别按图③、图④用它设计一个长方形桌面.请分别求出图③、图④中长方形的面积$y(m^{2})$与$DE$的长$x(m)$之间的函数表达式,并分别求出面积的最大值.

答案： 20.解:由$BC = 2\mathrm{m}$,$\angle C = 90^{\circ}$,$S_{\triangle ABC} = 1.5\mathrm{m}^2$,
得$AC = 1.5\mathrm{m}$,$AB = 2.5\mathrm{m}$.
(1)图①:$\because$ 四边形$CDEF$为正方形,
$\therefore CD = DE$,$\angle CDE = 90^{\circ}$.
$\therefore \angle ADE = 90^{\circ} = \angle C$.
$\because \angle DAE = \angle BAC$,$\therefore \triangle ADE \sim \triangle ACB$.$\therefore \frac{DE}{CB} = \frac{AD}{AC}$
即$\frac{DE}{2} = \frac{1.5 - DE}{1.5}$.$\therefore DE = \frac{6}{7}(\mathrm{m})$.
$\therefore S_{正方形DEFC} = \frac{6}{7} × \frac{6}{7} = \frac{36}{49}$.
图②:$\because$ 四边形$DEFG$为正方形,
$\therefore DE = DG$,$DE // AB$,$\angle DGB = \angle GDE = 90^{\circ}$.
$\therefore \triangle CDE \sim \triangle CAB$.$\therefore \frac{CD}{AC} = \frac{DE}{AB}$.$\therefore \frac{CD}{DE} = \frac{AC}{AB} = \frac{1.5}{2.5} = \frac{3}{5}$.
$\therefore$ 设$CD = 3a$,$DE = 5a$.
$\therefore DG = 5a$,$CE = \sqrt{DE^2 - CD^2} = 4a$.
$\because \angle ADG + \angle CDE = 90^{\circ}$,$\angle CDE + \angle CED = 90^{\circ}$,
$\therefore \angle ADG = \angle CED$.
$\because \angle AGD = 90^{\circ} = \angle C$,$\therefore \triangle AGD \sim \triangle DCE$.$\therefore \frac{AD}{DE} = \frac{DG}{CE}$.
即$\frac{1.5 - 3a}{5a} = \frac{5a}{4a}$.$\therefore a = \frac{6}{37}$.$\therefore DE = \frac{30}{37}$.
$\therefore S_{正方形DEFG} = \frac{30}{37} × \frac{30}{37} = \frac{900}{1369}$.
$\because \frac{36}{49} ＞ \frac{900}{1369}$,$\therefore$ 图①的正方形面积大.
(2)图③:由$Rt\triangle ADE \sim Rt\triangle ACB$,得$\frac{AD}{DE} = \frac{AC}{CB} = \frac{3}{4}$,
则$AD = \frac{3}{4}x$,$\therefore DC = AC - AD = \frac{6 - 3x}{4}$.
$\therefore y = DE × DC = x × \frac{6 - 3x}{4} = -\frac{3}{4}(x - 1)^2 + \frac{3}{4}$.
当$x = 1$时,长方形的面积有最大值为$\frac{3}{4}\mathrm{m}^2$.
图④,由$Rt\triangle DEC \sim \triangle ABC$,得$\frac{DE}{DC} = \frac{AB}{AC} = \frac{5}{3}$,
所以$DC = \frac{3}{5}x$,$\therefore DA = AC - DC = \frac{3}{2} - \frac{3}{5}x$.
由$Rt\triangle ADG \sim \triangle ABC$,得$\frac{DG}{DA} = \frac{BC}{BA} = \frac{4}{5}$,
则$DG = \frac{4}{5}DA = \frac{4}{5}\left(\frac{3}{2} - \frac{3}{5}x\right)$,
$\therefore y = DE × DG = x × \frac{4}{5}\left(\frac{3}{2} - \frac{3}{5}x\right) = -\frac{12}{25}\left(x - \frac{5}{4}\right)^2 + \frac{3}{4}$.
当$x = \frac{5}{4}$时,长方形的面积有最大值为$\frac{3}{4}\mathrm{m}^2$.