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# Area of Triangle in Circle

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Area of Triangle in Circle [#permalink]  04 Aug 2009, 19:34
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Question Stats:

88% (01:20) correct 13% (00:03) wrong based on 33 sessions
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) sqr(3)/4
B) 1/2
C) (pie)/4
D) 1
E) sqr(2)
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Re: Area of Triangle in Circle [#permalink]  04 Aug 2009, 22:48
Area of traingle = 1/2 (bh)

So we have to maximise the base and height. The base and height of inscribed triangle with one vertex at the center are maximum when both are equal to the radius = 1.

Therefore, the maximum area would be 1/2 (1*1) = 1/2
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Manager
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Re: Area of Triangle in Circle [#permalink]  05 Aug 2009, 01:30
robertrdzak wrote:
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) sqr(3)/4
B) 1/2
C) (pie)/4
D) 1
E) sqr(2)

What we can understand from the question stem is that the triangle is an Isoceles triangle.

So an isoceles triangle has max area when it is a rt. angled triangle

so max area =0.5 * b * h= 0.5 * r * r = 0.5 * 1 * 1= 0.5..hence B
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Re: Area of Triangle in Circle [#permalink]  05 Aug 2009, 20:59
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another way of looking at this can be area of the triangle A = 1/2 * radius*radius*sin (x).....area eqn interms of side and angle...

we can see that sine values are max only at x=90 degrees....hence we have A = 1/2 *1*1*1 = 1/2...OA B

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Re: Area of Triangle in Circle [#permalink]  25 Feb 2011, 23:51
scarish wrote:
Area of traingle = 1/2 (bh)

So we have to maximise the base and height. The base and height of inscribed triangle with one vertex at the center are maximum when both are equal to the radius = 1.

Therefore, the maximum area would be 1/2 (1*1) = 1/2

Although you got the correct answer, your explanation is incorrect. Since the triangle has a vertex at the center and the other two are on the circle with radius =1, at most all three sides can equal 1. BUT the area of a triangle is 1/2 *b*h. Thus you need to find h when the three sides are equal to 1. You will quickly realize your answer is no longer 1/2!

The correct way to reach 1/2 as an answer is to understand that an isosceles triangle has the max area when it is a right angled triangle. Now you know that the two sides are 1 and the angle between them at the center of the circle is 90. Thus the base or hypotenuse of the triangle will be 1/sqr2. This gives the area of 1/2 * b * h = 1/2 * 1/sqr2 * sqr2 = 1/2
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Re: Area of Triangle in Circle [#permalink]  26 Feb 2011, 02:03
Expert's post
This question is discussed here: maximum-area-no-clues-91398.html
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Re: Area of Triangle in Circle   [#permalink] 26 Feb 2011, 02:03
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