Question

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已知:如图,△ABC∽△DEF,其中∠B=∠E=90°,AB=DF=2a,DE=2EF 求:线段AC与DE的长（用含a的代数式表示）

Answer 1

∵⊿ABC∽⊿DEF,∠B=∠E
∴AC∶DF=AB∶DE,AC=DF×AB÷DE

∵DE²+EF²=DF²,∴(2EF)²+EF²=(2a)²
∴5EF²=4a²,EF=2√5a/5,DE=4√5a/5
∴AC=2a×2a÷(4√5a/5)=√5a

Answer 2

△ABC∽△DEF,
DE:EF=AB:BC
2EF:EF=2a:BC
2:1=2a:BC
BC=a,
AC²=AB²+BC²=4a²+a²=5a²,
AC=(√5)a;

DE:DF=AB:AC,
DE:2a=2a:(√5)a,
DE=4a/(√5)=(4√5)a/5;

Answer 3

DE=2EF
由勾股有,DE^2+EF^2=DF^2
则5EF^2=4a^2
EF=2√5a/5
则DE=2EF=4√5a/5

由相似得:AC:DF=AB:DE
则:AC=AB*DF/DE=√5a