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Geometry - 4 petals

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Geometry - 4 petals

by Brent@GMATPrepNow » Sat Jan 31, 2009 1:22 pm
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Please note that this is not an official GMAT question; it’s my attempt to create difficult (650+ level) GMAT-style questions for this forum.
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by truplayer256 » Sat Jan 31, 2009 1:43 pm
Is the answer A? I seperated the square into 4 sections such that the area of each section is 4. From this, I found the area of quarter of a circle with radius 2. pi(2)^(2)/4=pi. Then I did 4-pi to find the area of one of the unshaded regions outside of the petal. 4-[2(4-pi)]= the area of one petal, which is 2pi-4. Since there are 4 petals, I did 4(2pi-4)=8pi-16..
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by Brent@GMATPrepNow » Sat Jan 31, 2009 1:46 pm
truplayer256 wrote:Is the answer A?
A it is!

Here's my soln:


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by earth@work » Sat Jan 31, 2009 1:47 pm
Is it A
let O be center of bigger square(with side centers A,B,C&D) and let X be the vertex of square between A & B;
1/4 Area of circle with centre A = (pi *r^2)/4 =4pi/4=pi
area of sq AOBX = 2^2=4
Unshaded area BXO = 4-pi .... now we have total 8 such area in the bigger square = 8*(4-pi)
Area of bigger square =4^4=16
Shaded area =16-8*(4-pi) = 8pi-16
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by gaggleofgirls » Sat Jan 31, 2009 8:18 pm
Clearly the really tricky part of this is to figure out what-what is equal to come know fraction of the flower pedal.

The only shapes I could come up with to work were 1/4 of the area of a circle - the triangle formed by 1/2 the side of the square to the midpoint of the circle and back to the vertex of the square and this will equal the area of 1/2 of a flower pedal, so there are 8 of these in the shaded regions.

1/4 area of the circle = pi*r^2/4 - pi4/4 = pi
area of the small triangle = 2/2*2 = 2
area of 1/2 of a flower pedal = (pi-2)
8(pi-2) = 8Pi-16
Answer = A

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