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Two squares, each of side lengths 1 unit and having their [#permalink]
 14 Sep 2010, 07:27
Question Stats:
66% (03:34) correct
33% (05:05) wrong based on 3 sessions
Two squares, each of side lengths 1 unit and having their centres at O, are rotated with respect to each other to generate octagon ABCDEFGH, as shown in the figure above. If AB = 43/99, find the area of the octagon. Attachment: 
G86.PNG [ 5.44 KiB | Viewed 1431 times ]
a. 73/99 b. 86/99 c. 1287/9801 d. 76/99
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can someone chime in? This is something I really dont know how to even start at!....
thanks
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RaviChandra wrote: Two squares, each of side lengths 1 unit and having their centres at O, are rotated with respect to each other to generate octagon ABCDEFGH, as shown in the figure above. If AB = 43/99, find the area of the octagon. Attachment: G86.PNG a. 73/99 b. 86/99 c. 1287/9801 d. 76/99 First note that all the triangles will have the same dimensions. Let the legs of these triangles be x and y, also we know that the hypotenuse equals to \frac{43}{99}, so x^2+y^2=(\frac{43}{99})^2. If we take one side of a square we'll see that x+y+\frac{43}{99}=1 --> x+y=\frac{56}{99} --> square it --> x^2+2xy+y^2=(\frac{56}{99})^2, as from above x^2+y^2=(\frac{43}{99})^2, then: (\frac{43}{99})^2+2xy=(\frac{56}{99})^2 --> 2xy=(\frac{56}{99})^2-(\frac{43}{99})^2=(\frac{56}{99}-\frac{43}{99})(\frac{56}{99}+\frac{43}{99})=\frac{13}{99}*1=\frac{13}{99}; Now, the are of the octagon will be area of a square minus area of 4 triangles --> area=1-4*(\frac{1}{2}*xy)=1-2xy=1-\frac{13}{99}=\frac{86}{99}. Answer: B.
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thanku so much for the reply
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EASY WAY is as shown below
Connect AO and BO to get a AOB. We get totally 8 triangles like this.
Now. Lets calculate the AREA of AOB
AREA = .5*base*hiegt = 0.5*43/99 * 1/2 note: i/2 is the hieght as the length of a side of the given square is 1 and half of that is what is the hieght
hence AREA = 1/4 * 43/99
As told we have 8 similar triangles in this octagon.
Hence ANSWER = 8*1/4 * 43/99 = 86/99
Hope it is clear
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muralimba wrote: EASY WAY is as shown below
AREA = .5*base*hiegt = 0.5*43/99 * 1/2 note: i/2 is the hieght as the length of a side of the given square is 1 and half of that is what is the hieght
This doesnt looks like a straight line i dont think i can be 0.5 Attachment: 
G86.PNG [ 5.65 KiB | Viewed 1364 times ]
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RaviChandra wrote: muralimba wrote: EASY WAY is as shown below
AREA = .5*base*hiegt = 0.5*43/99 * 1/2 note: i/2 is the hieght as the length of a side of the given square is 1 and half of that is what is the hieght
This doesnt looks like a straight line i dont think i can be 0.5 Attachment: G86.PNG It is the length of the perpendicular from the center of the square to its side AB. It has to be half the length of the side, hence 1/2.
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great solutions, all.. this helps a lot.
thanks
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I dont understand why the solutions remove the area of the four triangles outside the stationary square? Should not the area of octagon add the area of those four triangles rather than subtract? Bunuel ,Shrouded?
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