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In the figure above, the area of the parallelogram is

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by Gmat_mission » Thu Sep 24, 2020 2:50 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 psarma » Thu Sep 24, 2020 3:20 pm
Area of parallelogram is base x height
To find the height, drop an altitude to the base. This would form a 30-60-90 triangle, so sides would be in the ratio 1: \(\sqrt{3}\) :2
Thus, \(\sqrt{3}\) :2 = h:8
h =4 \(\sqrt{3}\)

Area = 4 \(\sqrt{3}\) X 12
= 48 \(\sqrt{3}\)
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Re: In the figure above, the area of the parallelogram is

by Scott@TargetTestPrep » Thu Oct 01, 2020 9:55 am
Gmat_mission wrote: ↑
Thu Sep 24, 2020 2:50 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
Solution:

By drawing the altitude, we see that we have a 30-60-90 triangle nested in the parallelogram, with 8 as the hypotenuse and the altitude as the side opposite the 60-degree angle. Since the ratio of the sides of a 30-60-90 triangle is x : x√3 : 2x, the base of the triangle will be x = 4, the altitude will be x√3 = 4√3, and the hypotenuse will be 2x = 8.

Therefore, the area of the parallelogram is base x height = 12 x 4√3 = 48√3.

Answer: D

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