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2-to-1 rectangle with circle.JPG [ 19.83 KiB | Viewed 3603 times ]

In the diagram above, the sides of rectangle ABCD have a ratio AB:BC = 1:2, and the circle is tangent to three sides of the rectangle. If a point is chosen at random inside the rectangle, what is the probability that it is not inside the circle?

In the diagram above, the sides of rectangle ABCD have a ratio AB:BC = 1:2, and the circle is tangent to three sides of the rectangle. If a point is chosen at random inside the rectangle, what is the probability that it is not inside the circle? (A) \(\frac{4-{\pi}}{4}\) (B) \(\frac{4+{\pi}}{4}\) (C) \(\frac{4+{\pi}}{8}\) (D) \(\frac{8-{\pi}}{8}\) (E) \(\frac{8+{\pi}}{8}\) For a discussion of Geometric Probability, as well as a complete explanation of this particular question, see: http://magoosh.com/gmat/2013/geometric- ... -the-gmat/ Mike

Let the smaller side of square = 2x Larger side will be = 4x Radius of the circle will be = x Area of the Square = 8\(x\)2 Area of the Circle = \(\pi\)\(x\)2

the probability that the point is inside the circle = Area of the Circle/ Area of the Square = (\(\pi\)\(x\)2)/ (8\(x\)2) = \(\pi\)/8

the probability that the point is not inside the circle = 1 - the probability that the point is inside the circle = 1- \(\pi\)/8

Answer D

Hope it helps
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Re: In the diagram above, the sides of rectangle ABCD have a rat [#permalink]

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08 Feb 2013, 03:00

2

This post received KUDOS

mikemcgarry wrote:

Attachment:

2-to-1 rectangle with circle.JPG

In the diagram above, the sides of rectangle ABCD have a ratio AB:BC = 1:2, and the circle is tangent to three sides of the rectangle. If a point is chosen at random inside the rectangle, what is the probability that it is not inside the circle?

Re: In the diagram above, the sides of rectangle ABCD have a rat [#permalink]

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08 Feb 2013, 09:08

Assume the sides of the rectangle to be 2 and 4. Diameter of the circle=AB=2 Radius of the circle=1 Area of circle= {\pi} Area of the rectangle = 2*4=8 Area of the rectangle outside circle = 8-{\pi} So, probability= 8-{\pi}/8

Re: In the diagram above, the sides of rectangle ABCD have a rat [#permalink]

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09 Feb 2013, 18:58

Let us assume the sides of the rectangle be 1 and 2, so the area of the rectangle is 2 which implies that the circle is inscribed within a square whose area is 1.

Area of circle inscribed within a square is \(\frac{pi}{4}\) times the area of square = \(\frac{pi}{4}\)

Probability of point not inside the circle = 1 - probability of point inside the circle = 1 - \((pi/4)/2\) = 1 - \(\frac{pi}{8}\) = \(\frac{(8-pi)}{8}\)

It's funny. I was wondering this same thing, and I had to quote a response of Bunuel in which he used the \({\pi}\) symbol to see what it looked like in the html text.

Basically, you type {\pi} ----- (open curvy brackets)(backstroke)("pi")(close curvy brackets) ---- and then highlight that in the "math" delimiters ---- the m button, under the bold button in the rtf bar at the top of the editing window, does this. All math symbols need to be within the "math" delimiters.

Does this make sense?

Mike
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Mike McGarry Magoosh Test Prep

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Re: In the diagram above, the sides of rectangle ABCD have a rat [#permalink]

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Re: In the diagram above, the sides of rectangle ABCD have a rat [#permalink]

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08 Jun 2016, 11:16

Good question kudos given!

Probability = (Area of rectangle - Area of circle) / area of rectangle

b = breadth , l = length, r = radius. we know l = 2b so area of rectangle = l * b = 2b^2 . we can see the diameter of the circle = b hence radius = b/2 area will be π(b^2) / 4