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The circular base of an above ground swiming pool lies in a [#permalink]

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02 Nov 2006, 10:49

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The circular base of an above ground swiming pool lies in a level yard and just touches two straight sides of a fence at points A and B, as shown in the figure. point C is on the ground where the two sides of the fence meet. how far from the center of the pool's base is point A?

(1) The base has are 250 square feet. (2) The center of the base is 20 feet from point C

This problem sucks. What does "as shown in the figure" mean? Given that the distance from C to the center of the circle is 20 feet, I can tell you the radius of the circle to any arbitrary precision. So if you want it down to the nearest micron, I need only get measuring instruments up to the task. If this was physics class, I would say I have 1 (or 2) significant figure in statement 2 so I need only give you the radius to that many significant figures. I can do that with any number of measuring instruments.

You just need to find the radius of the circle. With the area, you can solve for R. Statement 2 adds nothing of value.

So what did you think of my post above? If I can tell you the radius to any precision you require, I can tell you the radius. Thus, statement 2 is sufficient (although I am sure that the person who wrote the problem wouldn't agree).

Re: The circular base of an above ground swiming pool lies in a [#permalink]

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12 Jan 2013, 11:38

Hello!

Potentially a stupid question... When I reviewed the answer in the GMAT review book, I was wondering why statement 2 is wrong. If we know that the distance from C to Q (q being the center of the circle) is 20 can't we infer that triangle CQA is a right triangle that is 12-16-20 and therefore we know the other sides?

Re: The circular base of an above ground swiming pool lies in a [#permalink]

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13 Jan 2013, 03:54

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Expert's post

alexpavlos wrote:

Hello!

Potentially a stupid question... When I reviewed the answer in the GMAT review book, I was wondering why statement 2 is wrong. If we know that the distance from C to Q (q being the center of the circle) is 20 can't we infer that triangle CQA is a right triangle that is 12-16-20 and therefore we know the other sides?

Or is this logic flawed?

Thanks! Alex

Good question. Your assumption is correct but conclusion is not.

You are right, CQA IS a right triangle, because the tangent line (CA) is always at the 90 degree angle (perpendicular) to the radius (QA) of a circle.

So, we have that CQ=hypotenuse=20. BUT, knowing that hypotenuse equals to 20 DOES NOT mean that the sides of the right triangle necessarily must be in the ratio of Pythagorean triple - 12:16:20. Or in other words: if \(x^2+y^2=20^2\) DOES NOT mean that \(x=12\) and \(y=16\). Certainly this is one of the possibilities but definitely not the only one. In fact \(x^2+y^2=20^2\) has infinitely many solutions for \(x\) and \(y\) and only one of them is \(x=12\) and \(y=16\).

For example: \(x=1\) and \(y=\sqrt{399}\) or \(x=2\) and \(y=\sqrt{396}\)...

Re: The circular base of an above ground swiming pool lies in a [#permalink]

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13 Jan 2013, 04:59

The distance between the center of base to point A - Radius of the circular pool.

A. Area = 250 square feet = 22/7 * r2..can find the radius. Sufficient B. Distance from center of pool to point C = 20 feet. Doesn't help to find the radius. Insufficient

Re: The circular base of an above ground swiming pool lies in a [#permalink]

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25 May 2014, 15:33

Bunuel wrote:

alexpavlos wrote:

Hello!

Potentially a stupid question... When I reviewed the answer in the GMAT review book, I was wondering why statement 2 is wrong. If we know that the distance from C to Q (q being the center of the circle) is 20 can't we infer that triangle CQA is a right triangle that is 12-16-20 and therefore we know the other sides?

Or is this logic flawed?

Thanks! Alex

Good question. Your assumption is correct but conclusion is not.

You are right, CQA IS a right triangle, because the tangent line (CA) is always at the 90 degree angle (perpendicular) to the radius (QA) of a circle.

So, we have that CQ=hypotenuse=20. BUT, knowing that hypotenuse equals to 20 DOES NOT mean that the sides of the right triangle necessarily must be in the ratio of Pythagorean triple - 12:16:20. Or in other words: if \(x^2+y^2=20^2\) DOES NOT mean that \(x=12\) and \(y=16\). Certainly this is one of the possibilities but definitely not the only one. In fact \(x^2+y^2=20^2\) has infinitely many solutions for \(x\) and \(y\) and only one of them is \(x=12\) and \(y=16\).

For example: \(x=1\) and \(y=\sqrt{399}\) or \(x=2\) and \(y=\sqrt{396}\)...

Hope it's clear.

Hi Bunuel,

I ran into a similar issue but took it even further -- I inferred that triangle QAC is a 30.60.90. Not really sure my reasoning behind it come to think of it but that's what I ended up with. A 30.60.90 and a side, therefore statement 2 was sufficient.

Cant we assume that angle C is cut in half by the two triangles and therefore will be 30?

Re: The circular base of an above ground swiming pool lies in a [#permalink]

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26 May 2014, 03:03

Expert's post

russ9 wrote:

Bunuel wrote:

alexpavlos wrote:

Hello!

Potentially a stupid question... When I reviewed the answer in the GMAT review book, I was wondering why statement 2 is wrong. If we know that the distance from C to Q (q being the center of the circle) is 20 can't we infer that triangle CQA is a right triangle that is 12-16-20 and therefore we know the other sides?

Or is this logic flawed?

Thanks! Alex

Good question. Your assumption is correct but conclusion is not.

You are right, CQA IS a right triangle, because the tangent line (CA) is always at the 90 degree angle (perpendicular) to the radius (QA) of a circle.

So, we have that CQ=hypotenuse=20. BUT, knowing that hypotenuse equals to 20 DOES NOT mean that the sides of the right triangle necessarily must be in the ratio of Pythagorean triple - 12:16:20. Or in other words: if \(x^2+y^2=20^2\) DOES NOT mean that \(x=12\) and \(y=16\). Certainly this is one of the possibilities but definitely not the only one. In fact \(x^2+y^2=20^2\) has infinitely many solutions for \(x\) and \(y\) and only one of them is \(x=12\) and \(y=16\).

For example: \(x=1\) and \(y=\sqrt{399}\) or \(x=2\) and \(y=\sqrt{396}\)...

Hope it's clear.

Hi Bunuel,

I ran into a similar issue but took it even further -- I inferred that triangle QAC is a 30.60.90. Not really sure my reasoning behind it come to think of it but that's what I ended up with. A 30.60.90 and a side, therefore statement 2 was sufficient.

Cant we assume that angle C is cut in half by the two triangles and therefore will be 30?

Yes, angle C will be cut in half by CQ. But we don't know the measure of angle C. So, we cannot assume that it's 60 degrees and that half of it is 30 degrees.

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