Question

在梯形ABCD中,AD‖BC,AC⊥DB,AC=5,∠DBC=30度.求BD的长和梯形ABCD的面积

Answer 1

解：

 ∵∠DBC=30,AD‖BC

 ∴∠ADB=30

 ∴BE=2CE,DE=2AE

BD=BE+DE=2CE+2AE=2（CE+AE）=2×5=10

S梯形ABCD=S△ABE+S△CBE+S△ADE+S△CDE

         =1/2×AE×BE+1/2×CE×BE+1/2×AE×DE+1/2×CE×DE

         =1/2×BE(AE+CE)+1/2×DE(AE+CE)

         =1/2×BE×AC+1/2×DE×AC

         =1/2×AC(BE+DE)

         =1/2×AC×BD

         =1/2×5×10

         =25

我是下午才看到的，虽然有点晚，还是希望你能采纳！0(∩_∩)0

Answer 2

BD=5乘根号下3，面积＝25/2乘根号下3