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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]  07 Apr 2012, 20:39
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Equilateral.png [ 1.79 KiB | Viewed 1054 times ]
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, 02:14, edited 1 time in total.
Edited the question and the diagram
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Re: Area of a Quadrilateral enclosed in a Triangle [#permalink]  07 Apr 2012, 23:08
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]  08 Apr 2012, 02:34
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boomtangboy wrote:
Attachment:
The attachment Equilateral.png is no longer available
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 1049 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] 08 Apr 2012, 02:34
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