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2-to-1 rectangle with circle.JPG [ 19.83 KiB | Viewed 2855 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.

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

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19 Aug 2015, 04:57

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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

Simplifying we get (8 - π)/8

gmatclubot

Re: In the diagram above, the sides of rectangle ABCD have a rat
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08 Jun 2016, 11:16

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