题目内容
分析:根据相似三角形的判定与性质得出
=
,
=
,进而求出QZ的长即可得出答案.
| CB |
| CZ |
| AB |
| QZ |
| FG |
| ZG |
| EF |
| QZ |
解答:
解:根据已知得出:
AB∥DZ∥EF,
∴
=
,
=
,
∵AB=1.5m,EF=1.7m,FG=1.7m,ZF=BZ+1.3,BC=1,
∴
=
,
=
,
∴QZ=1.5+BZ,
∴
=
,
解得:BZ=3,
∴QZ=6m.
故答案为:6.
AB∥DZ∥EF,
∴
| CB |
| CZ |
| AB |
| QZ |
| FG |
| ZG |
| EF |
| QZ |
∵AB=1.5m,EF=1.7m,FG=1.7m,ZF=BZ+1.3,BC=1,
∴
| 1 |
| 1+BZ |
| 1.5 |
| QZ |
| 1.7 |
| 1.7+1.3+BZ |
| 1.7 |
| QZ |
∴QZ=1.5+BZ,
∴
| 1.7 |
| 3+BZ |
| 1.7 |
| 1.5+BZ |
解得:BZ=3,
∴QZ=6m.
故答案为:6.
点评:此题主要考查了相似三角形的判定与性质,根据已知得出三角形对应边之间的关系是解题关键.
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