Question

已知,如图,平行四边形ABCD的两条对角线AC与BD相交于点O,直线m与AC相交于点p,做AC⊥m,CF⊥m

Answer 1

（1）S四边形ABCD=AC*BD/2=10*8/2=40,

（2）作DH⊥AC于H,

则∠ODH=30°,OH=OD/2=2,DH=2根号3,

∴S△ACD=AC*DH/2=10根号3,

∴S平行四边形ABCD=2S△ACD=20根号3；

（3）分别作BG⊥AC于G,DH⊥AC于H,

则BG=OB/Sinθ,DH=OD/Sinθ,

S四边形ABCD=S△ACB+S△ACD

=AC*BG/2+AC*DH/2

=AC(BG+DH)/2

=AC(OB*Sinθ+OD*Sinθ)/2

=AC*(OB+OD)*Sinθ/2

=AC*BD*Sinθ/2

=abSinθ/2