《2025年名校课堂九年级数学上册北师大版》

11. 如图,将等边三角形$ABC沿边AC上的高线BD平移到\triangle EFG$,阴影部分的面积记为$S$.若$\frac {BF}{FD}= \frac {1}{3}$,$S_{\triangle ABC}= 16$,则$S= $9.

答案： 9

12. (本课时T7变式)如图,在正方形$ABCD$中,$E为BC$的中点,连接$ED交AC于点F$,则$S_{\triangle DFC}:S_{四边形ABEF}$的值为 (B)

A. $\frac {1}{4}$
B. $\frac {2}{5}$
C. $\frac {7}{13}$
D. $\frac {3}{8}$

答案： B

13. 如图,在$\triangle ABC$中,点$D,E,F分别在边AB,AC,BC$上,连接$DE,EF$.已知四边形$BFED$是平行四边形,$\frac {DE}{BC}= \frac {1}{4}$.
(1)若$AB= 8$,求线段$AD$的长.2
(2)若$\triangle ADE的面积为1$,求$□ BFED$的面积.6

答案： 解:
(1)
∵ 四边形 BFED 是平行四边形, $\therefore DE // BF. \therefore DE // BC. \therefore ∠ADE = ∠ABC, ∠AED = ∠ACB. \therefore △ADE \backsim △ABC. \therefore \frac{AD}{AB} = \frac{DE}{BC} = \frac{1}{4}. \because AB = 8, \therefore AD = 2$.
(2) $\because △ADE \backsim △ABC, \therefore \frac{S_{△ADE}}{S_{△ABC}} = (\frac{DE}{BC})^2 = (\frac{1}{4})^2 = \frac{1}{16}. \because S_{△ADE} = 1, \therefore S_{△ABC} = 16. \because$ 四边形 BFED 是平行四边形, $\therefore EF // AB. \therefore ∠CEF = ∠CAB, ∠CFE = ∠CBA. \therefore △EFC \backsim △ABC. \therefore \frac{DE}{BC} = \frac{BF}{BC} = \frac{1}{4}, \therefore \frac{FC}{BC} = \frac{3}{4}. \therefore \frac{S_{△EFC}}{S_{△ABC}} = (\frac{FC}{BC})^2 = (\frac{3}{4})^2 = \frac{9}{16}. \therefore S_{△EFC} = 9. \therefore S_{//ogram BFED} = 16 - 9 - 1 = 6$.

14. 如图,在$\triangle ABC$中,$AB= 5$,$BC= 3$,$AC= 4$,$PQ// AB$,点$P在边AC$上(与点$A,C$不重合),点$Q在边BC$上.
(1)当$\triangle PQC的面积与四边形PABQ$的面积相等时,求$CP$的长.$2\sqrt{2}$
(2)当$\triangle PQC的周长与四边形PABQ$的周长相等时,求$CP$的长.$\frac{24}{7}$
(3)若在$AB上存在一点M$,使得$\triangle PQM$为等腰直角三角形,请直接写出$PQ$的长.$\frac{60}{37}$或$\frac{120}{49}$

答案： 解:
(1) $\because PQ // AB, \therefore △PQC \backsim △ABC. \therefore \frac{S_{△PQC}}{S_{△ABC}} = (\frac{CP}{CA})^2. \because S_{△PQC} = S_{四边形 PABQ}, \therefore S_{△PQC} : S_{△ABC} = 1 : 2. \therefore \frac{CP}{CA} = \sqrt{\frac{1}{2}} = \frac{\sqrt{2}}{2}. \therefore CP = \frac{\sqrt{2}}{2}CA = 2\sqrt{2}$.
(2) $\because △PQC \backsim △ABC, \therefore \frac{CP}{CA} = \frac{CQ}{CB} = \frac{PQ}{AB}. \because AC = 4, AB = 5, BC = 3, \therefore \frac{CP}{4} = \frac{CQ}{3} = \frac{PQ}{5}$. 设 $\frac{CP}{4} = \frac{CQ}{3} = \frac{PQ}{5} = k$, 则 $CP = 4k, CQ = 3k, PQ = 5k. \therefore C_{△PQC} = 12k$, $C_{四边形 PABQ} = PA + AB + BQ + PQ = 4 - CP + 5 + 3 - CQ + PQ = 4 - 4k + 5 + 3 - 3k + 5k = 12 - 2k. \because C_{△PQC} = C_{四边形 PABQ}, \therefore 12k = 12 - 2k$, 解得 $k = \frac{6}{7}. \therefore CP = 4k = \frac{24}{7}$.
(3) PQ 的长为 $\frac{60}{37}$ 或 $\frac{120}{49}$.