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In the figure above, the area of the parallelogram is
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22 Jul 2020, 02:15
22 Jul 2020, 02:15
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In the figure above, the area of the parallelogram is
A. 40
B. \(24\sqrt{3}\)
C. 72
D. \(48\sqrt{3}\)
E. 96
Re: In the figure above, the area of the parallelogram is
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22 Jul 2020, 02:16
22 Jul 2020, 02:16
Expert Reply
Post A Detailed Correct Solution For The Above Questions And Get A Kudos.
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
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GRE Prep Club - MATH Book for the GRE Exam
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
For more GRE Quant Theory and Practice, see
GRE Prep Club - MATH Book for the GRE Exam
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Re: In the figure above, the area of the parallelogram is
[#permalink]
22 Jul 2020, 07:15
22 Jul 2020, 07:15
1
Carcass wrote:
In the figure above, the area of the parallelogram is
A. 40
B. \(24\sqrt{3}\)
C. 72
D. \(48\sqrt{3}\)
E. 96
Area of parallelogram = (base)(height)
Start by drawing an extra line, which also happens to be the height of the parallelogram
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, . . .
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
Area of parallelogram = (base)(height)
= (12)(4√3)
= 48√3
Answer: D
