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triangles and circles.... the worst...

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by cneal4 » Mon May 04, 2009 7:42 am
can you post a figure I don't understand the question.
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Picuture of figure radius 1 and BC 1

by jhia » Mon May 04, 2009 8:59 am
Hi,

Please see the attached excel document to view the picture of the figure and provide explanation as to how they came to the answer : squarte root of 2/2?


thanks.
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picture.xlsx
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Correction on ANSWER

by jhia » Mon May 04, 2009 9:01 am
correction on the answer: the answer is square root of 3/2 (Not square root of 2/2 as noted in the PREVIOUS POST)
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by kic883 » Sat May 09, 2009 6:44 am
I think there is something missing of the graph.

Is angle "B" a right angle ?
If B is a right angle, we can find AB = sqrt(3),

detail please check here https://www.platinumgmat.com/gmat_study_ ... an_theorem

BC = 1
AB = sqrt(3)

BC * AB * 1/2 = sqrt(3)/2
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by helviolc » Tue May 19, 2009 11:43 am
kic883 wrote:I think there is something missing of the graph.

Is angle "B" a right angle ?
If B is a right angle, we can find AB = sqrt(3),

detail please check here https://www.platinumgmat.com/gmat_study_ ... an_theorem

BC = 1
AB = sqrt(3)

BC * AB * 1/2 = sqrt(3)/2

You dont need to know that B is a right angle to solve this problem.

From center O, trace a segment to B, forming triangle OBC. Since this segment starts from O and ends in the circumference, you know this distance OB is equal to the radius = 1.

OC is also equal to 1. From the exercise, you know BC is equal to 1. Therefore, triangle OBC is equilateral.

The height of this equilateral triangle must be equal to the height of triangle in question ABC - from B trace a perpendicular line to segment AC to visualize this better.

The height of an equilateral triangle is sqrt(3) / 2 * L (to see this, apply the Pythagorean theorem to any equilateral triangle to compare the height and the side). Since L is equal to 1, the height is sqrt(3) / 2.

The base of the triangle is segment AC = 2. Do the math: (Base x Height) / 2 and you will have the answer sqrt(3) / 2 - the second answer from top.

Hope this helps.

H
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An Alternate Method

by rsadana1 » Tue May 19, 2009 5:06 pm
From the information in the question as I understand:
Radius = 1
AC is diameter = 2 (Since Radius = 1)

Now since Angle ABC is inscribed by diameter of the circle Angle ABC = 90
Thus, we can apply pythagoras theorem on Triangle ABC
one leg BC = 1
Hypotenuse AC = 2

Thus from the theorem AB = sqrt (3)
Area of Triangle ABC = 1/2 * height * base = 1/2 * 1 * sqrt(3)

Final Answer = sqrt(3)/2
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