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7. 如图，工地上竖立着两根电线杆$AB$、$CD$。现分别自两杆上的点$A$、$C$处向两侧地面上的点$E$、$D$和点$B$、$F$引钢丝绳并拉紧，以固定电线杆。已知$AB = 4\mathrm{m}$，$CD = 6\mathrm{m}$，求钢丝绳$AD$与$BC$的交点$P$离地面的高度。
答案
解：过点P作PG⊥EF ，垂足为点G ，如图所示
由题意得， AB//CD//PG
因为AB//PG
所以△PGD∽△ABD
所以$\frac {PG}{AB}=\frac {DG}{DB}$
因为CD//PG
所以△PGB∽△CDB，
所以$\frac {PG}{CD}=\frac {GB}{DB}$
因为$\frac {DG}{DB}+\frac {GB}{DB}=\frac {DB}{DB}=1$
所以$\frac {PG}{AB}+\frac {PG}{CD}=1$
因为AB=4m，CD=6m
所以$\frac {PG}{4}+\frac {PG}{6}=1$
所以$PG=\frac {12}{5}m$
答：交点P离地面的高度为$\frac {12}{5}m$
1. 长度如下的四条线段中，成比例的是()
A. $1 \mathrm { cm }, 2 \mathrm { cm }, 3 \mathrm { cm }, 4 \mathrm { cm }$
B. $1.5 \mathrm { cm }, 2.5 \mathrm { cm }, 4.5 \mathrm { cm }, 6.5 \mathrm { cm }$
C. $\sqrt { 2 } \mathrm { cm }, \sqrt { 3 } \mathrm { cm }, \sqrt { 5 } \mathrm { cm }, \frac { \sqrt { 30 } } { 2 } \mathrm { cm }$
D. $3 \mathrm { cm }, 4 \mathrm { cm }, 5 \mathrm { cm }, 6 \mathrm { cm }$
A. $1 \mathrm { cm }, 2 \mathrm { cm }, 3 \mathrm { cm }, 4 \mathrm { cm }$
B. $1.5 \mathrm { cm }, 2.5 \mathrm { cm }, 4.5 \mathrm { cm }, 6.5 \mathrm { cm }$
C. $\sqrt { 2 } \mathrm { cm }, \sqrt { 3 } \mathrm { cm }, \sqrt { 5 } \mathrm { cm }, \frac { \sqrt { 30 } } { 2 } \mathrm { cm }$
D. $3 \mathrm { cm }, 4 \mathrm { cm }, 5 \mathrm { cm }, 6 \mathrm { cm }$
答案
C
2. 在下列所给的条件中，能判定$\triangle A B C \backsim \triangle A ^ { \prime } B ^ { \prime } C ^ { \prime }$的是()
A. $A B = \frac { 3 } { 2 }, B C = 6, A ^ { \prime } B ^ { \prime } = 16, B ^ { \prime } C ^ { \prime } = 12, \angle A = \angle A ^ { \prime }$
B. $\angle A = 70 ^ { \circ }, \angle B = 35 ^ { \circ }, \angle A ^ { \prime } = 70 ^ { \circ }, \angle C ^ { \prime } = 85 ^ { \circ }$
C. $A B = 4, B C = 6, A C = 8, A ^ { \prime } B ^ { \prime } = 12, B ^ { \prime } C ^ { \prime } = 18, A ^ { \prime } C ^ { \prime } = 24$
D. $\angle C = \angle C ^ { \prime } = 90 ^ { \circ }, A B = 15, A C = 5, A ^ { \prime } B ^ { \prime } = 5, B ^ { \prime } C ^ { \prime } = \frac { 5 } { 3 }$
A. $A B = \frac { 3 } { 2 }, B C = 6, A ^ { \prime } B ^ { \prime } = 16, B ^ { \prime } C ^ { \prime } = 12, \angle A = \angle A ^ { \prime }$
B. $\angle A = 70 ^ { \circ }, \angle B = 35 ^ { \circ }, \angle A ^ { \prime } = 70 ^ { \circ }, \angle C ^ { \prime } = 85 ^ { \circ }$
C. $A B = 4, B C = 6, A C = 8, A ^ { \prime } B ^ { \prime } = 12, B ^ { \prime } C ^ { \prime } = 18, A ^ { \prime } C ^ { \prime } = 24$
D. $\angle C = \angle C ^ { \prime } = 90 ^ { \circ }, A B = 15, A C = 5, A ^ { \prime } B ^ { \prime } = 5, B ^ { \prime } C ^ { \prime } = \frac { 5 } { 3 }$
答案
C
3. 如图，$D E // B C, E F // A B$，则()
A. $\frac { A D } { D B } = \frac { D E } { B C }$
B. $\frac { A D } { D B } = \frac { D E } { F C }$
C. $\frac { A E } { E C } = \frac { D E } { B C }$
D. $\frac { E F } { A B } = \frac { C F } { F B }$
(https://p3-hippo-sign.byteimg.com/tos-cn-i-a9yeduch1e/5c8c8c8c8c8c4c8c9c8c8c8c8c8c8c8c~tplv-ii0cwwkcx9-resize-crop-v1:600:470:737:550:0:0.png?lk3s=f89d68d8&x-expires=2045275247&x-signature=7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d)
A. $\frac { A D } { D B } = \frac { D E } { B C }$
B. $\frac { A D } { D B } = \frac { D E } { F C }$
C. $\frac { A E } { E C } = \frac { D E } { B C }$
D. $\frac { E F } { A B } = \frac { C F } { F B }$
(https://p3-hippo-sign.byteimg.com/tos-cn-i-a9yeduch1e/5c8c8c8c8c8c4c8c9c8c8c8c8c8c8c8c~tplv-ii0cwwkcx9-resize-crop-v1:600:470:737:550:0:0.png?lk3s=f89d68d8&x-expires=2045275247&x-signature=7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d)
答案
B
