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If points A and C both lie on the circle with center B and

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If points A and C both lie on the circle with center B and [#permalink] New post 09 Dec 2012, 00:46
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At the end of the explanation for this question, they state what would happen if the information was different, just to explain for interest's sake.

But I cannot understand how you can know that if AC is 2 1/2 times larger that the triangle is isosceles.

Read till the end:


Image here: platinumgmat dot com/global/images/inline_questions/00024-1.gif

If points A and C both lie on the circle with center B and the measurement of angle ABC is not a multiple of 30, what is the ratio of the area of the circle centered at point B to the area of triangle ABC?
A) 2π
B) 2π(AB)2/(BC)2
C) 4π
D) π(BC)2/.5(BC)(AB)
E) None of the Above

Begin by finding the area of the circle:
Areacircle = πr2
Areacircle = π(AB)2 = π(BC)2
In dealing with triangle ABC, BC = AB since both are radii. At this point, some students make a mistake and assume that AB is the height of the triangle and BC is the base of the triangle (or vice versa). However, we cannot assume that BC is the base and AB is the height since we have not yet shown that ABC is a right triangle. You could only make BC the base and AB the height if triangle ABC were a right triangle (in which case AB would be a perpendicular segment drawn from a vertex, A, to the side opposite that vertex, B).
By definition, the height of a triangle is the length of a segment drawn from a vertex perpendicular to the side opposite that vertex. A line that is perpendicular to the side opposite a vertex will, by definition, form a 90 degree angle. Consequently, for line AB to be the height of triangle ABC, angle ABC must be a right angle (i.e., 90 degrees).
Since the question states that "the measurement of angle ABC is not a multiple of 30," angle ABC cannot be 30, 60, 90, 120, etc. Consequently, angle ABC is not a right angle and line AB is not the height of triangle ABC.
Without the height, you cannot determine the area of the triangle. Without the area of the triangle, you do not have enough information to solve the problem. The correct answer is It Cannot Be Determined.

Then they say this:
Note: If the question omitted the words "the measurement of angle ABC is not a multiple of 30" and instead said that the length of line AC is 2 1/2 times larger than the radius, you would be dealing with a 45-45-90 right triangle with sides r, r, and r*2 1/2. In this instance with a right triangle, the area of the triangle would be (1/2)bh = (1/2)(r)(r) = .5r2 and the ratio of the area of the circle centered at point B to the area of triangle ABC would be 2π.

Can anyone shed light on this?
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Re: AC is 2 1/2 times larger than the radius [#permalink] New post 11 Dec 2012, 06:09
jessello wrote:
At the end of the explanation for this question, they state what would happen if the information was different, just to explain for interest's sake.

But I cannot understand how you can know that if AC is 2 1/2 times larger that the triangle is isosceles.

Read till the end:


Image here: platinumgmat dot com/global/images/inline_questions/00024-1.gif

If points A and C both lie on the circle with center B and the measurement of angle ABC is not a multiple of 30, what is the ratio of the area of the circle centered at point B to the area of triangle ABC?
A) 2π
B) 2π(AB)2/(BC)2
C) 4π
D) π(BC)2/.5(BC)(AB)
E) None of the Above

Then they say this:
Note: If the question omitted the words "the measurement of angle ABC is not a multiple of 30" and instead said that the length of line AC is 2 1/2 times larger than the radius, you would be dealing with a 45-45-90 right triangle with sides r, r, and r*2 1/2. In this instance with a right triangle, the area of the triangle would be (1/2)bh = (1/2)(r)(r) = .5r2 and the ratio of the area of the circle centered at point B to the area of triangle ABC would be 2π.

Can anyone shed light on this?



A couple of things:
1.) If you post your specific questions on the GMAT quant P.S forum, you will get a better response.
2.) There is an option to upload an image. Please use that instead of using URLs.


Answer to your question:

The triangle will always be isosceles because AB=BC=radius. (unless AC=AB=AC is given; then it will be an equilateral triangle)

Also, triangle will be a 45-45-90 when AC= (2r)^(1/2) and not two and a half times larger than radius. Try to apply the Pythagoras theorem on triangle ABC.

Does that help?
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Re: AC is 2 1/2 times larger than the radius [#permalink] New post 11 Dec 2012, 13:06
Hi jumsumtak

Quote:
A couple of things:
1.) If you post your specific questions on the GMAT quant P.S forum, you will get a better response.
2.) There is an option to upload an image. Please use that instead of using URLs.


I could not post URLs or IMGs until I had posted 5 times. Not my fault.
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Re: AC is 2 1/2 times larger than the radius [#permalink] New post 11 Dec 2012, 13:08
I'm talking about this statement:

Quote:
If the question omitted the words "the measurement of angle ABC is not a multiple of 30" and instead said that the length of line AC is 2 1/2 times larger than the radius, you would be dealing with a 45-45-90 right triangle


How do they know that?
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Re: AC is 2 1/2 times larger than the radius [#permalink] New post 12 Dec 2012, 02:59
jessello wrote:
I'm talking about this statement:

Quote:
If the question omitted the words "the measurement of angle ABC is not a multiple of 30" and instead said that the length of line AC is 2 1/2 times larger than the radius, you would be dealing with a 45-45-90 right triangle


How do they know that?



1.)
In general:
If you know the relation between 3 sides (AB:BC:AC::x:y:z) then you can calculate the angles of the triangle.

2.)
In this case: 2 sides are equal AB=BC=r and the length of the third side is given. Hence, you can calculate the angles of the triangle

3.)
Either you have copied the explanation incorrectly or the source is wrong.

As I explained earlier if AC= (2r)^(1/2) then it will be a 45-45-90 triangle.

we know, AB=BC=r and AC= (2r)^(1/2); apply AB^2+BC^2=AC^2 (Pythagoras theorem) and you will find that this equation holds true for triangle ABC. This in turn means the triangle is right angled at angle ABC.
Re: AC is 2 1/2 times larger than the radius   [#permalink] 12 Dec 2012, 02:59
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