题目内容
如图,小明同学用自制的直角三角形纸板DEF测量树的高度AB,他调整自己的位置,设法使斜边DF保持水平,并且边DE与点B在同一直线上.已知纸板的两条直角边DF=50cm,EF=30cm,测得边DF离地面的高度AC=1.5m,CD=20m,则树高AB为( )
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A.12m | B.13.5m | C.15m | D.16.5m |
D.
试题分析:∵∠DEF=∠BCD=90°,∠D=∠D,∴△DEF∽△DCB. ∴
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∵DF=50cm=0.5m,EF=30cm=0.3m,AC=1.5m,CD=20m,∴由勾股定理求得DE=0.40m.
∴
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故选D.
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