If each side of ΔACD above has length 3 and if AB has length : GMAT Problem Solving (PS)
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# If each side of ΔACD above has length 3 and if AB has length

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If each side of ΔACD above has length 3 and if AB has length [#permalink]

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07 Apr 2012, 19:39
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If each side of ΔACD above has length 3 and if AB has length 1, what is the area of region BCDE?

(A) $$\frac{9}{4}$$

(B) $$\frac{7}{4} \sqrt{3}$$

(C) $$\frac{9}{4} \sqrt{3}$$

(D) $$\frac{7}{2} \sqrt{3}$$

(E) $$6 + \sqrt{3}$$
[Reveal] Spoiler: OA

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Last edited by Bunuel on 08 Apr 2012, 01:14, edited 1 time in total.
Edited the question and the diagram
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07 Apr 2012, 22:08
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equilateral triangle hence Angle A = 60 deg
thus, in Triangle ABE, AB = 1 BE = 3^(1/2) and AE = 2

Area BCDE = Area ACD - Area ABE

Area ACD = 3^(1/2)/4 * (side)^2

Area ABE = 1/2 * 3^(1/2) * 1

we get 3.5/2 * 3^(1/2) multiplying numerator and denominator by 2 we get B
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Re: If each side of ΔACD above has length 3 and if AB has length [#permalink]

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08 Apr 2012, 01:34
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boomtangboy wrote:
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If each side of ΔACD above has length 3 and if AB has length 1, what is the area of region BCDE?

(A) $$\frac{9}{4}$$

(B) $$\frac{7}{4} \sqrt{3}$$

(C) $$\frac{9}{4} \sqrt{3}$$

(D) $$\frac{7}{2} \sqrt{3}$$

(E) $$6 + \sqrt{3}$$

Attachment:

Equilateral.png [ 1.79 KiB | Viewed 4699 times ]

Since each side of ΔACD has length 3 then ACD is an equilateral triangle and its each angle is 60°.

Now, the are of equilateral triangle is $$area_{equilateral}=side^2*\frac{\sqrt{3}}{4}=9\frac{\sqrt{3}}{4}$$ (for more check math-triangles-87197.html);

Next, since angle A is 60° then right triangle ABE is a 30°-60°-90° right triangle. Now, in a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio $$1 : \sqrt{3}: 2$$, the leg opposite 30° (AB) corresponds with $$1$$ and the leg opposite 60° (BE) corresponds with $$\sqrt{3}$$ (the hypotenuse AE corresponds with 2). So, since $$AB=1$$ then $$BE=\sqrt{3}$$, and the area of ABE is $$\frac{1}{2}*AB*BE=\frac{\sqrt{3}}{2}$$;

The area of of region BCDE is the area of ACD minus the area of ABE: $$9\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{2}=9\frac{\sqrt{3}}{4}-2\frac{\sqrt{3}}{4}=7\frac{\sqrt{3}}{4}$$.

Hope it's clear.
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Re: If each side of ΔACD above has length 3 and if AB has length [#permalink]

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14 Nov 2013, 07:29
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Re: If each side of ΔACD above has length 3 and if AB has length [#permalink]

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26 Apr 2016, 01:53
Hello from the GMAT Club BumpBot!

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Re: If each side of ΔACD above has length 3 and if AB has length   [#permalink] 26 Apr 2016, 01:53
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