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# Let P and Q be points which are two inches apart, and let A be the are

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Math Expert
Joined: 02 Sep 2009
Posts: 58453
Let P and Q be points which are two inches apart, and let A be the are  [#permalink]

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26 Nov 2018, 00:13
00:00

Difficulty:

75% (hard)

Question Stats:

43% (01:44) correct 57% (01:30) wrong based on 159 sessions

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Let P and Q be points which are two inches apart, and let A be the area, in square inches, of a circle which passes through P and Q. Which of the following is the set of all possible values of A.

A. 0 < A

B. 0 < A <= π

C. A = π

D. A > π

E. A ≥ π

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Re: Let P and Q be points which are two inches apart, and let A be the are  [#permalink]

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26 Nov 2018, 23:09
Bunuel wrote:
Let P and Q be points which are two inches apart, and let A be the area, in square inches, of a circle which passes through P and Q. Which of the following is the set of all possible values of A.

A. 0 < A

B. 0 < A <= π

C. A = π

D. A > π

E. A ≥ π

GMATinsight : can circle pass through any point on line P Q ? is so then shouldnt answer be B ..?
Manager
Joined: 14 Jun 2018
Posts: 217
Re: Let P and Q be points which are two inches apart, and let A be the are  [#permalink]

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30 Nov 2018, 00:18
Minimum area of the circle will be when P and Q are the two points of the diameter.
Area in that case would be π.

Max area will be when P and Q are the two point of the chord of a circle.
Diameter will be >2 inch , therefore , area > π

Manager
Joined: 16 Jul 2018
Posts: 59
Re: Let P and Q be points which are two inches apart, and let A be the are  [#permalink]

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28 Dec 2018, 06:00
Bunuel Hi could you please enlighten us with a solution for this problem

I thought that the distance between the two endpoints of a diameter are the lengthiest points in a circle, thus max case would be for r=1 thus A= π and all the other possible values should be 0<A<=π
Have I misunderstood the problem?
Manager
Joined: 02 Aug 2015
Posts: 153
Re: Let P and Q be points which are two inches apart, and let A be the are  [#permalink]

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28 Dec 2018, 07:50
1
UNSTOPPABLE12 wrote:
Bunuel Hi could you please enlighten us with a solution for this problem

I thought that the distance between the two endpoints of a diameter are the lengthiest points in a circle, thus max case would be for r=1 thus A= π and all the other possible values should be 0<A<=π
Have I misunderstood the problem?

Hey UNSTOPPABLE12,

If the given points are not the points of diameter, then the diameter will be even longer, making the area greater.

In other words, if a chord that is not the diameter of a circle is x, then the diameter should be surely greater than x. So the area will be greater than pi in such cases.

PS: If you are still unable to visualize the concept, I suggest that you draw two rough circles; one with given points as diameter and other with given points are chords, and see how area increases.

Cheers!
Manager
Joined: 16 Jul 2018
Posts: 59
Let P and Q be points which are two inches apart, and let A be the are  [#permalink]

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28 Dec 2018, 08:07
Diwakar003
thanks for the prompt response yes indeed i falsely thought of the max scenario to be that this 2 inches is the diameter, excluding in this way a variety of solutions, whereas by taking into consideration that this 2 inches is just a chord we can choose choice E A>=π , which covers even the possibility of these two points being the diameter.
Let P and Q be points which are two inches apart, and let A be the are   [#permalink] 28 Dec 2018, 08:07
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