如图,在平行四边形ABCD中,AE⊥BC,AF⊥CD,垂足分别为E、F.若AE=3cm,AF=4cm,AD=8cm,求C

1﹚由于CD=6,则平行四边形ABCD的周长=2﹙AD＋CD﹚=2﹙8＋6﹚=28
2﹚∵AE⊥BC而AD∥BC ∴∠EAD=90° 又∵∠EAF=60° ∴∠DAF=30°
∴在RT⊿AFD中 ∠D=90°－∠DAF=90°－30°=60°
∴在平行四边形ABCD中 ∠B=∠D=60° ∠BAD=∠BCD=180°－∠D=120°
3﹚∵平行四边形ABCD的周长为40cm ∴BC＋CD=20cm
又∵S平行四边形ABCD=AE·BC=AF·CD且AE=4cm,AF=6cm
∴4BC=6CD即2BC－3CD=0
∴得到方程组BC＋CD=20
2BC－3CD=0
解得BC=12 cm,CD=8cm ∴S平行四边形ABCD=BC·AE=48cm²
4﹚∵BC=12cm,AD与BC之间的距离为5cm
∴S平行四边形ABCD=12×5=60cm²
而S⊿BOC=¼S平行四边形ABCD=15cm²

1﹚根据你的求法：应该CD=6，则平行四边形ABCD的周长=2﹙AD＋CD﹚=2﹙8＋6﹚=28
2﹚∵AE⊥BC而AD∥BC ∴∠EAD=90° 又∵∠EAF=60° ∴∠DAF=30°
∴在RT⊿AFD中 ∠D=90°－∠DAF=90°－30°=60°
∴在平行四边形ABCD中 ∠B=∠D=60° ∠BAD=∠BCD=180°－∠D=120...

Intermezzo Flexi-offices
In analyzing flexi-offices ti became clear to us that there is a real,daily tension in the effectuation of their usage,This tension between the intended,traditional ,stati...

1﹚根据你的求法：应该CD=6，则平行四边形ABCD的周长=2﹙AD＋CD﹚=2﹙8＋6﹚=28
2﹚∵AE⊥BC而AD∥BC ∴∠EAD=90° 又∵∠EAF=60° ∴∠DAF=30°
∴在RT⊿AFD中 ∠D=90°－∠DAF=90°－30°=60°
∴在平行四边形ABCD中 ∠B=∠D=60° ∠BAD=∠BCD=180°－∠D=120...