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Find the area of AEDC, if ABCDEF is a regular hexagon with each side b

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Find the area of AEDC, if ABCDEF is a regular hexagon with each side b  [#permalink]

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New post 01 Jun 2017, 00:30
3
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A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

56% (02:53) correct 44% (02:55) wrong based on 84 sessions

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


A. \(\frac{√3a^2}{4}\)
B. \(\frac{3√3a^2}{4}\)
C. \(√3a^2\)
D. \(\frac{5√3a^2}{4}\)
E. \(\frac{7√3a^2}{4}\)
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Find the area of AEDC, if ABCDEF is a regular hexagon with each side b  [#permalink]

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New post 01 Jun 2017, 01:16
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
mugeYL1.png [ 10.08 KiB | Viewed 4841 times ]


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 OCD 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.
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Re: Find the area of AEDC, if ABCDEF is a regular hexagon with each side b  [#permalink]

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New post 01 Jun 2017, 09:01
nguyendinhtuong wrote:
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
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Find the area of AEDC, if ABCDEF is a regular hexagon with each side b  [#permalink]

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New post 01 Jun 2017, 15:44
niteshwaghray wrote:
Find the area of AEDC, if ABCDEF is a regular hexagon with each side being equal to “a” units.
Image


A. \(\frac{√3a^2}{4}\)
B. \(\frac{3√3a^2}{4}\)
C. \(√3a^2\)
D. \(\frac{5√3a^2}{4}\)
E. \(\frac{7√3a^2}{4}\)


I used sample numbers to solve:

1) Split up the hexagon into 6 equilateral triangles (something you should know for the exam)

2) Each side has length "a" so I replaced "a" with an easy number "2"

3) Find the area of one of the equilateral triangles with side "2" (a formula you should know) which ends up giving an area of \(\sqrt{3}\)

4) Realize that the AEDC contains exactly 4 equilateral triangles each with area \(\sqrt{3}\) .. There's a few ways to realize how many equilateral triangles there are... but i think it's safe to assume that the GMAT wouldn't give you a weird shape to find the area

5) Answer = 4 * \(\sqrt{3}\)
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Re: Find the area of AEDC, if ABCDEF is a regular hexagon with each side b  [#permalink]

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New post 01 Jun 2017, 17:02
Hello
Could you tell me how you figured out that there were 4 equilateral triangles in AEDC? thanks
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Re: Find the area of AEDC, if ABCDEF is a regular hexagon with each side b  [#permalink]

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New post 02 Jun 2017, 10:24
Adricht wrote:
Hello
Could you tell me how you figured out that there were 4 equilateral triangles in AEDC? thanks


See the attached picture.
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Re: Find the area of AEDC, if ABCDEF is a regular hexagon with each side b  [#permalink]

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New post 02 Jun 2017, 12:21
Attachment:
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Used some trigonometric functions such as

Area of a triangle can be written as = 1/2 side1 * side 2 * angle containing side 1 * side 2
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Re: Find the area of AEDC, if ABCDEF is a regular hexagon with each side b  [#permalink]

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New post 10 Nov 2017, 06:11
If you are aware of trigonometric concepts, you can solve this sum within a minute.

Area of a triangle = 1/2 * product of two sides * Sin(angle between them)
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Re: Find the area of AEDC, if ABCDEF is a regular hexagon with each side b &nbs [#permalink] 10 Nov 2017, 06:11
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