Question

正方形ABCD的边长为2,点E、F分别在BC、CD,且CE=CF,△AEF的面积等于1,求EF的长.

Answer 1

设CE=CF＝x

则

BE=DF=2-x

所以

S△AEF=SABCD-S△ABE-S△ADF-S△EFC

2*2-(1/2)*2*(2-x)-(1/2)*2*(2-x)-(1/2)*x*x＝1

4-(2-x)-(2-x)-(1/2)x^2=1

(1/2)x^2-2x+1=0

x^2-4x+4=2

(x-2)^2=2

x=2-√2

EF^2=EC^2+FC^2

=(2-√2)^2+(2-√2)^2

=2(2-√2)^2

EF=√2(2-√2)

=2√2-2