在平行四边形ABCD中,AC.BD相交于点O,若平行四边形ABCD的周长为64,且△AOB的周长比△BOC的周长多8,则
2个回答

因为ABCD是平行四边形,所以

AB = DC

BC = AD

OA = OC

四边形ABCD的周长为64,即

AB + DC + BC + AD = 64

AB + BC = 32

△AOB的周长 = AB + BO + OA

△BOC的周长 = BC + BO + OC

△AOB的周长比△BOC的周长多8,即

(AB + BO + OA) - (BC + BO + OC) = 8

AB - BC = 8

联立

AB + BC = 32

AB - BC = 8

AB = 20

BC = 12