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Re: If each side of ΔACD above has length 3 and if AB has length
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08 Apr 2012, 01:34

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boomtangboy wrote:

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

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Equilateral.png [ 1.79 KiB | Viewed 7217 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: \(9\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{2}=9\frac{\sqrt{3}}{4}-2\frac{\sqrt{3}}{4}=7\frac{\sqrt{3}}{4}\).