Let ABCDEF be a regular hexagon. Let P be the midpoint of BC, and let Q be the midpoint of EF. Find the ratio of the area of APDQ to the area of ABCDEF.

Guest May 13, 2020

#2**+1 **

See the following image

Let the side of the hexagon = S

Then the area of the hexagon = 6(1/2)S^2 sqrt (3) / 2 = 3sqrt(3)S^2 / 2 = (3/2) sqrt (3)S^2

Note that the area of triangle ABP =

(1/2)(AB)(BP) sin ABP =

(1/2)(S)(1/2)S * sin (120) =

(1/2) (S) (1/2)S * sqrt (3) / 2 =

(1/8)sqrt (3) S^2

Using symmetry we have three other triangles congruent to triangle ABP

So....the total area of these 4 triangles = (1/2)sqrt(3) S^2

So...the area of APDQ = (3/2 - 1/2)sqrt(3)S^2 = sqrt (3)S^2

So.....the ratio of the area of APDQ to the area of the hexagon =

sqrt (3)S^2 2

_____________ = ___

(3/2) sqrt (3) S^2 3

CPhill May 13, 2020