niteshwaghray wrote:

Find the area of AEDC, if ABCDEF is a regular hexagon with each side being equal to “a” units.

A. \(\frac{√3a^2}{4}\)

B. \(\frac{3√3a^2}{4}\)

C. \(√3a^2\)

D. \(\frac{5√3a^2}{4}\)

E. \(\frac{7√3a^2}{4}\)

Attachment:

mugeYL1.png

Each angle of ABCDEF is 120 degree.

O is the center of triangle ACE. We have \(\widehat{AOE} =\widehat{COE} = \widehat{AOC} = 120^o\)

We could easily have \(\Delta AFE = \Delta EDC = \Delta CBA = \Delta AOC = \Delta COE = \Delta EOA\)

Hence \(S_{ AFE} = S_{ EDC} = S_{ CBA} = S_{AOC} = S_{COE} = S_{EOA}\)

OD cuts EC at M. Triangle OEC is an equilateral triangle. Hence \(S_{OCD}=\frac{1}{2} \times a \times \frac{a\sqrt{3}}{2}=\frac{a^2\sqrt{3}}{4}\)

Hence \(S_{OEDC}=2S_{OCD}=2 \times \frac{a^2\sqrt{3}}{4} = \frac{a^2\sqrt{3}}{2}\)

So \(S_{OEC}=\frac{1}{2} S_{OEDC} =\frac{1}{2} \times \frac{a^2\sqrt{3}}{2} = \frac{a^2\sqrt{3}}{4}\)

\(S_{AEDC}=4S_{OEC}=4 \times \frac{a^2\sqrt{3}}{4}=a^2\sqrt{3}\)

The answer is C.

May I suggest another approach (probably not precise as yours, but it also worked in this case):

The area of a regular hexagon is (3sqrt{3}a^2)/2.

We see that there are 6 equal triangles (AOE=AFE=EOC=EDC...) and that 4 of them are part of ACDE. Therefore I suggest we have to multiply the area of the hexagon by 4/6 or 2/3: ((3sqrt{3}a^2)/2)*2/3=sqrt{3}a^2