Question

已知,如图,等腰梯形ABCD中,AD//BC,对角线AC垂直于BD且相交于点P,AD=3,BC=7,

Answer 1

由于ABCD是等腰梯形,易知

ABC全等于DCB

角BAC=角CDB

从而

ACP全等于DCP

从而

AP=DP

BP=CP

解得

AP=DP=3sqrt(2)/2

BP=CP=7sqrt(2)/2

ABCD面积=四个三角形面积之和

=1/2(AP*DP+DP*CP+CP*BP+BP*AP)

==1/2(AP+CP)(DP+BP)

1/2*AC*BD=1/2*(10srqt(2)/2)*(10sqrt(2)/2)

=25