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3. 如图,在$△ ABC$中,$AB=AC=5$,$BC=6$,$AD⊥ BC$,垂足为$D$,点$P$是边$AB$上的一个动点,过点$P$作$PF// AC$交线段$BD$于点$F$,作$PG⊥ AB$交$AD$于点$E$,交线段$CD$于点$G$,设$BP=x$.
(1)用含$x$的代数式表示线段$DG$的长;
(2)设$△ DEF$的面积为$y$,求$y$与$x$之间的函数关系式,并写出自变量的取值范围;
(3)$△ PEF$能否为直角三角形?如果能,求出$BP$的长;如果不能,请说明理由.
(1)用含$x$的代数式表示线段$DG$的长;
(2)设$△ DEF$的面积为$y$,求$y$与$x$之间的函数关系式,并写出自变量的取值范围;
(3)$△ PEF$能否为直角三角形?如果能,求出$BP$的长;如果不能,请说明理由.
答案
3. (1) $\because AB=AC=5,BC=6, AD⊥BC$ 于点 $D, \therefore BD=\frac{1}{2}BC=\frac{1}{2}×6=3. \because △ABD∽△GBP, \therefore \frac{PB}{BG}=\frac{BD}{AB}, \therefore \frac{x}{BG}=\frac{3}{5}, \therefore BG=\frac{5}{3}x. \therefore DG=\frac{5}{3}x-3.$
(2) $\because ∠GED=∠B, \therefore \frac{DE}{DG}=\frac{BD}{AD}=\frac{3}{4}, \therefore DE=\frac{3}{4}(\frac{5}{3}x-3)=\frac{5}{4}x-\frac{9}{4}. \because PF// AC, \therefore \frac{BF}{BC}=\frac{BP}{AB}, \therefore \frac{BF}{6}=\frac{x}{5}. \therefore BF=\frac{6}{5}x, \therefore DF=3-\frac{6}{5}x. \therefore y=\frac{1}{2}(\frac{5}{4}x-\frac{9}{4})(3-\frac{6}{5}x)=-\frac{3}{4}x^2+\frac{129}{40}x-\frac{27}{8}. \therefore y=-\frac{3}{4}x^2+\frac{129}{40}x-\frac{27}{8}, 自变量的取值范围: \frac{9}{5}＜x＜\frac{5}{2}.$
(3) 若 $EF⊥PG$ 时,
$\because EF⊥PG,ED⊥FG, \therefore ∠FED+∠DEG=90°,∠FED+∠EFD=90°, \therefore ∠FED + ∠DEG = 90°, ∠FED+∠EFD=90°, \therefore ∠EFD=∠DEG, 且 ∠EDF=∠EDG, \therefore △EDF ∽ △GDE. \therefore \frac{ED}{DG}=\frac{DF}{ED}. \therefore ED^2=FD · DG. 代入数据,解得 x=\frac{9}{5}(不合题意,舍去),x=\frac{125}{57}.$
若 $EF⊥PF, \therefore ∠PFB+∠EFD=90°, 且 ∠PFB=∠ACB,∠ACB+∠DAC=90°. \therefore ∠EFD=∠DAC, 且 ∠EDF=∠ADC=90°. \therefore △EDF∽△CDA. \therefore \frac{ED}{DF}=\frac{CD}{AD}, 即 \frac{\frac{5}{4}x-\frac{9}{4}}{3-\frac{6}{5}x}=\frac{3}{4}. 解得 x=\frac{90}{43}.$
综上所述,$△PEF$ 能为直角三角形,$BP=\frac{125}{57}或\frac{90}{43}.$
(2) $\because ∠GED=∠B, \therefore \frac{DE}{DG}=\frac{BD}{AD}=\frac{3}{4}, \therefore DE=\frac{3}{4}(\frac{5}{3}x-3)=\frac{5}{4}x-\frac{9}{4}. \because PF// AC, \therefore \frac{BF}{BC}=\frac{BP}{AB}, \therefore \frac{BF}{6}=\frac{x}{5}. \therefore BF=\frac{6}{5}x, \therefore DF=3-\frac{6}{5}x. \therefore y=\frac{1}{2}(\frac{5}{4}x-\frac{9}{4})(3-\frac{6}{5}x)=-\frac{3}{4}x^2+\frac{129}{40}x-\frac{27}{8}. \therefore y=-\frac{3}{4}x^2+\frac{129}{40}x-\frac{27}{8}, 自变量的取值范围: \frac{9}{5}＜x＜\frac{5}{2}.$
(3) 若 $EF⊥PG$ 时,
$\because EF⊥PG,ED⊥FG, \therefore ∠FED+∠DEG=90°,∠FED+∠EFD=90°, \therefore ∠FED + ∠DEG = 90°, ∠FED+∠EFD=90°, \therefore ∠EFD=∠DEG, 且 ∠EDF=∠EDG, \therefore △EDF ∽ △GDE. \therefore \frac{ED}{DG}=\frac{DF}{ED}. \therefore ED^2=FD · DG. 代入数据,解得 x=\frac{9}{5}(不合题意,舍去),x=\frac{125}{57}.$
若 $EF⊥PF, \therefore ∠PFB+∠EFD=90°, 且 ∠PFB=∠ACB,∠ACB+∠DAC=90°. \therefore ∠EFD=∠DAC, 且 ∠EDF=∠ADC=90°. \therefore △EDF∽△CDA. \therefore \frac{ED}{DF}=\frac{CD}{AD}, 即 \frac{\frac{5}{4}x-\frac{9}{4}}{3-\frac{6}{5}x}=\frac{3}{4}. 解得 x=\frac{90}{43}.$
综上所述,$△PEF$ 能为直角三角形,$BP=\frac{125}{57}或\frac{90}{43}.$
4. 如图，AB和A'B'与x轴垂直，点A的坐标是(2,4)，△AOB和△A'OB'是位似三角形，且位似比是1：3，点C是OA'的中点，反比例函数$y=\frac{k}{x}(x＞0)$的图像经过点C，与A'B'交于点D.
(1)求点D的坐标；
(2)连接BD、CD，求四边形ABDC的面积.
(1)求点D的坐标；
(2)连接BD、CD，求四边形ABDC的面积.
答案
4. (1)$\because △AOB$ 和 $△A'OB'$ 是位似三角形,且位似比是 $1:3, \therefore \frac{AB}{A'B'}=\frac{OB}{OB'}=\frac{1}{3}.$
$\because 点 A 的坐标是(2,4), \therefore OB=2,AB=4, \therefore \frac{4}{A'B'}=\frac{2}{OB'}=\frac{1}{3}, \therefore A'B'=12,OB'=6, \therefore A'(6,12). \because 点 C 是 OA' 的中点, \therefore C(3,6), \therefore k=3×6=18, 即反比例函数的表达式为 y=\frac{18}{x}.$
$\because A'B'⊥x 轴, \therefore x_D=x_{A'}=6, \therefore y_D=\frac{18}{6}=3, 即 D(6,3).$
(2) 如图.
$\because A(2,4)、B(2,0)、C(3,6)、D(6,3)、A'(6,12)、B'(6,0), \therefore S_{△OAB}=\frac{1}{2}×2×4=4, S_{△DBB'}=\frac{1}{2}×(6-2)×3=6, S_{△A'OB'}=\frac{1}{2}×6×12=36, S_{△A'CD}=\frac{1}{2}×(12-3)×(6-3)=\frac{27}{2}, \therefore S_{四边形ABDC}=S_{△OA'B'}-S_{△DBB'}-S_{△OAB}-S_{△A'CD}=36-6-4-\frac{27}{2}=\frac{25}{2}.$
