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In the figure above, the area of the parallelogram is

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by Vincen » Sat Dec 04, 2021 7:41 am
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2019-04-26_1721.png
In the figure above, the area of the parallelogram is

A. \(40\)
B. \(24\sqrt3\)
C. \(72\)
D. \(48\sqrt3\)
E. \(96\)

Answer: D

Source: Official Guide
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Re: In the figure above, the area of the parallelogram is

by Brent@GMATPrepNow » Sun Dec 05, 2021 6:25 am
Vincen wrote: ↑
Sat Dec 04, 2021 7:41 am
 2019-04-26_1721.png

In the figure above, the area of the parallelogram is

A. \(40\)
B. \(24\sqrt3\)
C. \(72\)
D. \(48\sqrt3\)
E. \(96\)

Answer: D

Source: Official Guide
Area of parallelogram = (base)(height)

Start by drawing an extra line, which also happens to be the height of the parallelogram
Image
This creates a special 30-60-90 right triangle


When we compare the blue 30-60-90 right triangle with the purple BASE 30-60-90 right triangle, . . .
Image
We see that the blue 30-60-90 right triangle is 4 times bigger than the purple BASE 30-60-90 right triangle, . .
So, the missing lengths are 4 and 4√3

At this point, we know the base and the height
Image

Area of parallelogram = (base)(height)
= (12)(4√3)
= 48√3

Answer: D
Brent Hanneson - Creator of GMATPrepNow.com
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