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What is the area of the figure below?
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08 Dec 2018, 07:38
08 Dec 2018, 07:38
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What is the area of the figure below?
#GRE practice question what is the area of the figure below.jpg [ 10.41 KiB | Viewed 4598 times ]
A. \(96 \sqrt{3}\)
B. \(256\)
C. \(256 \sqrt{3}\)
D. \(512\)
E. It cannot be determined by the information given.
Attachment:
#GRE practice question what is the area of the figure below.jpg [ 10.41 KiB | Viewed 4598 times ]
A. \(96 \sqrt{3}\)
B. \(256\)
C. \(256 \sqrt{3}\)
D. \(512\)
E. It cannot be determined by the information given.
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Re: What is the area of the figure below?
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10 Dec 2018, 08:19
10 Dec 2018, 08:19
1
Carcass wrote:
What is the area of the figure below?
A. \(96 \sqrt{3}\)
B. \(256\)
C. \(256 \sqrt{3}\)
D. \(512\)
E. It cannot be determined by the information given.
Attachment:
#GRE practice question what is the area of the figure below.jpg
A. \(96 \sqrt{3}\)
B. \(256\)
C. \(256 \sqrt{3}\)
D. \(512\)
E. It cannot be determined by the information given.
Useful rule: the sum of the angles in an n-sided polygon = (n - 2)(180°)
Here, we have a 6-sided figure (hexagon), so the sum of the angles = (6 - 2)(180°) = (4)(180°) = 720°
Since all 6 angles are EQUAL, the measurement of each angle = 720°/6 = 120°
So, we HAVE:
Now place a point in the CENTER and draw lines to each vertex.
We get:
Since all 6 angles in the center must add to 360°, each answer must equal 60°
At this point, we can see that the 6-sided figure is comprised of 6 EQUILATERAL triangles.
Area of an equilateral triangle = (side²/4)√3
The area of ONE equilateral triangle = (8²/4)√3
= (64/4)√3
= 16√3
So, the area of ALL 6 triangles = (6)(16√3) = 96√3
Answer: A
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Re: What is the area of the figure below?
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15 Jan 2023, 09:54
15 Jan 2023, 09:54
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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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