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 Post subject: GMAT Geometry
PostPosted: Mon Apr 08, 2013 10:35 am 
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In the figure above, lines L and P are parallel. The segment AD is the same length as the segment DC; AD is parallel to BC. If the length of AD is equal to 4, and angle ADC is equal to 60º, what is the area of ABCD?

A. 4√2
B. 4√3
C. 8√2
D. 8√3
E. 16√2

(D) AB is parallel to DC and AD is parallel to BC, so ABCD is parallelogram. We know how to find the area of a parallelogram we multiply the height of parallelogram by its base length.

We are told that the angle ADC is a 60º angle. If we draw an altitude from A straight down and perpendicular to line P, the length of that altitude will be equal to the height of the rhombus. Moreover, it forms a 60-30-90 triangle, so we can easily find its length. Since the hypotenuse of that triangle is equal to 4, the second longest side must be equal to 2√3 (since the proportions for the triangle run x, x√3, and 2x).

To find the area of a rhombus, multiply the height of the rhombus by its base length: 4 × 2√3 = 8√3, or answer choice (A).
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The sum of the interior angles should be (n – 2) × 180. So the sum of the interior angles for ABCD = 360⁰. How is it determined that the traingle formed from A is a 30-60-90 triangle?
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 Post subject: Re: GMAT Geometry
PostPosted: Mon Apr 08, 2013 10:36 am 
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We determine that triangle EDA is a 90-60-30 triangle because:

angle ADC is 60 degrees (we know that from the question statement)
angle AED is 90 degrees because AE is height of parallelogram ABCD
angle EAD is 30 degrees since it equals 180⁰ – angle AED - angle ADC = 180⁰ – 90⁰ – 60⁰ = 30⁰
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 Post subject: Re: GMAT Geometry
PostPosted: Mon Apr 08, 2013 10:47 am 
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I think the answer should be 16√3. As area of each triangle of the parallelogram would be 1/2 × 4 × 4 × sin(60⁰) = 8√3.


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 Post subject: Re: GMAT Geometry
PostPosted: Mon Apr 08, 2013 10:49 am 
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questioner wrote:
I think the answer should be 16√3. As area of each triangle of the parallelogram would be 1/2 × 4 × 4 × sin(60⁰) = 8√3.
sin(60⁰) = √3/2.


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