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DS: Geometry

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Director
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DS: Geometry [#permalink] New post 07 Dec 2004, 15:57
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
DS from Kaplan
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Intern
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 [#permalink] New post 07 Dec 2004, 16:06
I think it should be D.
from both the stems, you could figure out the value of side of a square, and the angle for the arc is 45, so there you go.
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 [#permalink] New post 07 Dec 2004, 16:20
foraj wrote:
I think it should be D.
from both the stems, you could figure out the value of side of a square, and the angle for the arc is 45, so there you go.


Pls explain in itsy bitsy bits! Ta very much!
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 [#permalink] New post 07 Dec 2004, 16:48
AB is the diameter of the circle, and ABCD is a square, these are given in the question.
Stem 1) r=5, so you know the side of square is 10, the area of triangle ADC will be 1/2*10*10, shaded region is area of ADC - area of Arc AT(T is where AC intersects with the circle). Area of Arc is 45/360*2*3.14*5, now you have both.
Stem 2) Similarly as above.
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Re: DS: Geometry [#permalink] New post 07 Dec 2004, 17:05
gayathri wrote:
DS from Kaplan


I'd say D, too.

I think this is hard to very hard.

In order to do it, the plan would be to find the area of the triange, and subtract the little arc piece from it.

Once we have the radius, which is four, we know everything about the square, which has side 8, and the triangle would be 32.

But the question is still about the arc.

For that, we see that the angle of CAD is 45. Therefore, if we drew a line OT (T being where AC hits the circle) we'd have an arc, and the degrees would be double CAD, so AOT is 90 degrees.

If that's true, then we know everything. The area of the whole wedge from AOT is 1/4 the area of the circle (because the arc angle is 90) and we can figure out the area of the triangle AOT because it's a 45 45 90 triangle with side 4.

So the area of the little arc we're subtracting is the area of the whole wedge minus the area of the triangle. Since we can get that, we can get the whole thing.

Same goes for two. With the triangle area, we can work backwards to everything else.
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 [#permalink] New post 07 Dec 2004, 17:59
OA is D. Great explanation Ian!
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 [#permalink] New post 07 Dec 2004, 18:02
is it A.
A. is sufficient we can find the area of square , triangle and the sector.
B. can give diff values for the base of the triangle and is not sufficient.
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 [#permalink] New post 07 Dec 2004, 18:14
batliwala wrote:
is it A.
A. is sufficient we can find the area of square , triangle and the sector.
B. can give diff values for the base of the triangle and is not sufficient.


How do you get different values for the base?
ABCD is a square so AD = DC = x
So if area of ADC = 32
=> x^2/2 =32; x^2 = 64
so x = 8, since x cannot be -8.

=> radius of circle is 4. Same info as A.
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 [#permalink] New post 08 Dec 2004, 00:10
:wall , yup agree costly mistake.
  [#permalink] 08 Dec 2004, 00:10
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