Question

在梯形ABCD中,E在AD上,F在BC上,CD//EF//AB,CD=2.AB=4.EF将梯形ABCD的面积两等分,求E

Answer 1

设梯形AEFB 高为h1 梯形CDEF高为h2 梯形ABCD高为h

S梯形CDEF=（CD+EF）h2/2 =梯形ABCD/2=(AB+CD)h/4 ∴h2=3h/(EF+2)(1)

S梯形AEFB=（AB+EF）h1/2 =梯形ABCD/2=(AB+CD)h/4 ∴h1=3h/(EF+4)(2)(1)+(2)得 h1+h2=3h/(EF+2)+3h/(EF+4) 即h=3h/(EF+2)+3h/(EF+4)

同除以h得 1=3/(EF+2)+3/(EF+4) EF=根号10