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

Question

Find the area of AEDC, if ABCDEF is a regular hexagon with each side being equal to “a” units.
 https://i.imgur.com/mugeYL1.png

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}

broall · Answer

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 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.

malevg · Answer

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

papagorgio · Answer

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}

Adricht · Answer

Hello 
Could you tell me how you figured out that there were 4 equilateral triangles in AEDC? thanks

papagorgio · Answer

See the attached picture.

Leo8 · Answer

FullSizeRender (9).jpg 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

pushkarajnjadhav · Answer

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)

Fdambro294 · Answer

Concept 1: In a Regular Hexagon, you can draw 6 Equilateral Triangles of Side = a from the Center to each Vertex.
Thus, the Area of a Regular Hexagon = 6 * (Area of an Equilateral Triangle of Side a)

6 *   = (3 * a^2 * sqrt3) / 2 = Area of ENTIRE Regular Hexagon

Concept 2: the Height of an Equilateral Triangle = a * sqrt3 / 2

if you Connect Vertex E to Vertex C, you will have an Equilateral Triangle in the Middle of the Regular Hexagon.

Side EC = Perpendicular Distance between the 2 Sides of the Regular Hexagon (Side FE to Side BC) = 2 * Altitude of any 1 of the 6 Equilateral Triangles that the Regular Hexagon can be divided into

Altitude of any 1 of the 6 Equilateral Triangles that the Regular Hexagon can be divided into = a * sqrt3 / 2

2 of these = 2 * a * sqrt3 / 2 = a * sqrt3

This means the Equilateral Triangle we formed in the middle by connecting Vertex E to Vertex C has Each Side = a *sqrt3

Concept 3: Since this is an Equilateral Triangle that we formed, this also means that Angle EAC = 60 Degrees

Further, by the S-A-S Congruency Rule, the 2 Triangles created by the Diagonals AE and AC are CONGRUENT Isosceles Triangles with the 2 Equal Sides = a

The Entire Angle at Vertex A = 120 degrees

Subtract out the 60 Degrees that is Angle EAC

and you have 60 degrees divided equally among Angle FAE and Angle BAC ------ Each Angle = 30 Degrees

Concept 4: Drawing a Perpendicular Height from the Apex Vertex (the Vertex b/w the 2 Equal Sides of the Isosceles Triangle) to the Non-Equal Side for Both of the Isosceles Triangles created by the 2 Diagonals----

this Perpendicular Height = Median = Angle Bisector

By drawing this Perpendicular Altitude in each of the Side Isosceles Triangles, you create 4 Congruent Right Triangles that are: 30-60-90 Right Triangles

Since the Side across from the 90 degree Angle = a

Leg 1 = a/2
Leg 2 = a*sqrt3 / 2

Concept 5: Area of Entire Regular Hexagon - 4 * (Area of 1 of the 30/60/90 Triangles) = Area of AEDC

Finally, take the Area of all 4 of these Triangles and Subtract it from the Regular Hexagon Area

(3 * a^2 * sqrt3) / 2 - 4 *   = Area of AEDC in the middle of the Hexagon

(3 * a^2 * sqrt3) / 2 - (a^2 * sqrt3) / 2 =

(2 * a^2 * sqrt3) / 2 =

a^2 * sqrt3

Answer C

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

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GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards