Question

如图a,在RT△ABC和RT△DEF中,∠ACB=∠DFE=90°,AC=6cm,BC=8cm,EF=5cm,DF=12

Answer 1

⑴将ΔAFG绕F逆时针旋转180°到ΔFBG‘,连接 HG’,

∵∠ACB=90°,∴∠A+∠B=90°,∴∠HBG‘=90°,

∴HG’^2=BH^2+BG‘^2=BH^2+AG^2,

∵∠DFE=90°,∴∠AFG+∠BFH=90°,∴∠HFG‘=90°,

∵FG=FG’,FH=FH,∴ΔFHG≌ΔFHG‘,∴GH=HG’,

∴AG^2+BH^2=GH^2.

⑵DE=√(EF^2DF^2)=13,AB=√(AC^2+BC^2)=10,F为AB的中点,∴AF=5,

过P作PQ⊥DF于Q,则PQ∥EF,∴ΔDPQ∽ΔDEF,

∴PQ/EF=DP/DE,AP=2X,∴PQ=10X/13,

∴SΔADP=1/2AD*PQ=1/2*7*10X/13=35X/13,

∴Y=SΔDEF-SΔADP=30-35X/13.

说明：本题应用旋转思想题目不是很难,难在量大.