Question

四边形abcd中,角BAD＝90,AB＝BC＝2倍根号3,AC=6,AD=3,求CD的长

Answer 1

过点B作BE⊥AC于E,过点D作DF⊥AC于F

∵AB＝BC＝2√3,AC＝6,BE⊥AC

∴AE＝CE＝AC/2＝3

∴cos∠BAC＝AE/AB＝3/2√3＝√3/2

∴∠BAC＝30

∵∠BAD＝90

∴∠CAD＝∠BAD-∠BAC＝60

∵AD＝3,DF⊥AC

∴AF＝AD/2＝3/2,DF＝3×√3/2＝3√3/2

∴CF＝AC-AF＝6-3/2＝9/2

∴CD²＝CF²+DF²＝81/4+27/4＝27

∴CD＝3√3