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A circle is inscribed in an isosceles trapezoid with bases 8

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Manager
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A circle is inscribed in an isosceles trapezoid with bases 8 [#permalink] New post 09 Oct 2004, 05:31
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A
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C
D
E

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A circle is inscribed in an isosceles trapezoid with bases 8 and 18. What is the area of the circle?

A. 36pi
B. 49pi
C. 64pi
D. 81pi
Manager
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 [#permalink] New post 09 Oct 2004, 06:22
the answer is A

anyway, I will explain as much as I could

AB and CD are the bases of trapezoid (AB=8 , CD=18)
draw the segment joining the midpoint of AB (M) to midpoint of CD (N)
The center should be the midpoint of MN. I called O.

The circle intersect BC at P
MB=BP=4
PC=CN=9 this implies BC=4+9=13

Notice that MN is a diameter of circle
After dropping the altitude from B to NC, we can know that the altitude is 12.
MN=12 implies radius is 6.

Hope I made it clear but I frankly could not see it as a GMAT question. Took me around 5 minutes.
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 [#permalink] New post 10 Oct 2004, 00:07
Great solution amernassar.
Thanks for the explanation.
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 [#permalink] New post 10 Oct 2004, 07:06
I do not understand how it is possible to find any solution to this pb. We just know bases lengths height of the trapezoid can be 10 or 3000...
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 [#permalink] New post 10 Oct 2004, 08:08
The problem says that a circle is inscribed in an isosceles trapezoid which means that there is a restriction of some kind on the trapezoid in order to be true, he is just asking in another way what height would it be for the circle to be inscribed.

Of course a height of 3000 for the isosceles trapezoid could not be circumscribed in any circle,just imagine it.


Another problem let's say. find the area of rectangle of width 5 inscribed in a circle of radius 6.5. Do we say that we have missing information for the length to solve it, or the restriction given to the problem solves the length of the rectangle and thus the problem.


Hope this twixt has helped in some way.
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 [#permalink] New post 10 Oct 2004, 09:02
twixt, i think that this is what you're missing:

From symmetry reasons, a perpendicular from the middle of base AB to the middle of base DC will divide the circle to two equal pieces.
The crucial thing to understand inorder to solve this question, is that MB, which is half of AB and equals to 4, is also equal to BP, where P is the point where the circle touches BC.
Two tangents to a circle from the same point are equal.
Once you find this, the question is easily solved.

I agree that this question seems out of scope!
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 [#permalink] New post 10 Oct 2004, 10:11
Equally important is to realize that an altitude from B to NC should be drawn, parallel and equal to the diameter, to form a triangle...then it becomes sort of easy. The difficulty lies in knowing what 'extra' work we have to do the diagram to solve the problem out - in other words, what the stem offers you may be good but not enough!
  [#permalink] 10 Oct 2004, 10:11
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A circle is inscribed in an isosceles trapezoid with bases 8

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