Let ABCDEF be a regular hexagon of unit side length. Line BE is extended to a point P outside of the hexagon such that FP=√3. The line AP intersects the lines BF and EF at points M and N, respectively. Calculate the area of quadrilateral BENM.
Since BF=FP=sqrt(3), triangle BFP is isosceles and angleFBP=angleFPB=30
angleFAE=30
angleAEP=anglePFA=150
AEPF is a parallelogram, so its diagonals bisect and FN=NE=1/2
also EP=1 so BP=3
AFPB is a trapezoid so its diagonals make triangles AMF and PMB similar with size ratio 3:1.
Since FB=sqrt(3), FM=sqrt(3)/4
[BENM] = [BFE] - [FMN]
=1/2*sqrt(3)*1 - 1/2*sqrt(3)/4*1/2
=7sqrt(3)/16
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Posted by Jer
on 2024-12-04 14:37:20 |