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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
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What is the greatest possible area of a triangular region with one
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22 Nov 2021, 06:52
22 Nov 2021, 06:52
Expert Reply
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Question Stats:
50% (02:37) correct
50% (02:03) wrong based on 4 sessions
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What is the greatest possible area of a triangular region with one vertex at the center of a circle of radius 1 and the other two vertices on the circle?
A. \(\frac{\sqrt{3}}{4}\)
B. \(\frac{1}{2}\)
C. \(\frac{\pi}{4}\)
D. 1
E. \(\sqrt{2}\)
A. \(\frac{\sqrt{3}}{4}\)
B. \(\frac{1}{2}\)
C. \(\frac{\pi}{4}\)
D. 1
E. \(\sqrt{2}\)
Re: What is the greatest possible area of a triangular region with one
[#permalink]
24 Nov 2021, 10:20
24 Nov 2021, 10:20
Expert Reply
Clearly two sides of the triangle will be equal to the radius of 1.
Now, fix one of the sides horizontally and consider it to be the base of the triangle.
\(area=\frac{1}{2}*base*height=\frac{1}{2}*1*height=\frac{height}{2}\).
So, to maximize the area we need to maximize the height. If you visualize it, you'll see that the height will be maximized when it's also equals to the radius thus coincides with the second side (just rotate the other side to see). which means to maximize the area we should have the right triangle with right angle at the center.
\(area=\frac{1}{2}*1*1=\frac{1}{2}\).
Answer: B.
Now, fix one of the sides horizontally and consider it to be the base of the triangle.
\(area=\frac{1}{2}*base*height=\frac{1}{2}*1*height=\frac{height}{2}\).
So, to maximize the area we need to maximize the height. If you visualize it, you'll see that the height will be maximized when it's also equals to the radius thus coincides with the second side (just rotate the other side to see). which means to maximize the area we should have the right triangle with right angle at the center.
\(area=\frac{1}{2}*1*1=\frac{1}{2}\).
Answer: B.
gmatclubot
Re: What is the greatest possible area of a triangular region with one [#permalink]
24 Nov 2021, 10:20
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