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A castle is at the center of the several flat paths which su
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08 Aug 2009, 09:19
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A castle is at the center of the several flat paths which surround it: 4 straight paths that travel from the castle to its circular moat, where they meet up with a perfectly circular path which borders the moat; that circular path circumscribes a square path which has its corners at the ends of the 4 straight paths—see the diagram to the right. If the total length of all of the pathways is q kilometers, then which expression represents distance from the castle to the circular moat? A) q / 4(2+2*sqrt(2)+pi) km B) q / 2(2+2*sqrt(2)+pi) km C) q / (2+2*sqrt(2)+pi) km D) 2q / (2+2*sqrt(2)+pi) km E) 4q / (2+2*sqrt(2)+pi) km
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Re: castle paths
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22 May 2014, 00:36
Let's take distance from castle to circular moat is r. side of square = sqrt2 * r Perimeter of Circle = 2* Pi * r Attachment: File comment: Castle Paths
gmatcastle.jpg [ 32.43 KiB  Viewed 5968 times ]
Total length of paths=q = 4* r + 4 *sqrt2 *r +2*pi*r q = (4+4*sqrt2 + 2* pi) r or , r = q/(4+4*sqrt2 + 2* pi) or, r = q/2(2+2*sqrt2+Pi) Hence, Answer is B
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Re: castle paths
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08 Aug 2009, 17:49
i couldnt figure out how to find the perimeter of the square in terms of r...



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Re: castle paths
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08 Aug 2009, 18:08
OA Blet distance from the castle to the circular moat is x distance from 4 straight paths that travel from the castle to its circular moat, where they meet up with a perfectly circular path which borders the moat ( addition of lenght of diagonals of sqr)= 4x Thus side of inscribed sqr = x*sqrt2 (use pythagoras 2x^ = (side of sqr)^2.....hence perimeter of sqr = 4*sqrt2*x circumference of perimeter = pi * (2*x) Thus our eqn becomes => q= pi * (2*x) + 4*sqrt2*x + 4x on simplification we get x = q / 2(2+2*sqrt(2)+pi) km... OA B
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Re: castle paths
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08 Aug 2009, 18:55
Thanks for the solution...



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Castle and the path 
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18 Nov 2009, 05:54
A castle is at the center of the several flat paths which surround it: 4 straight paths that travel from the castle to its circular moat, where they meet up with a perfectly circular path which borders the moat; that circular path circumscribes a square path which has its corners at the ends of the 4 straight paths. If the total length of all of the pathways is q kilometers, then which expression represents distance from the castle to the circular moat? 1) q/4(2+2\sqrt{2}+pie) 2) q/2(2+2\sqrt{2}+pie) 3) q/(2+2\sqrt{2}+pie) 4) 2q/(2+2\sqrt{2}+pie) 5) 4q/2(2+2\sqrt{2}+pie)
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Re: castle paths
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18 Nov 2009, 06:28
bhushan252 wrote: OA B
let distance from the castle to the circular moat is x
distance from 4 straight paths that travel from the castle to its circular moat, where they meet up with a perfectly circular path which borders the moat ( addition of lenght of diagonals of sqr)= 4x
Thus side of inscribed sqr = x*sqrt2 (use pythagoras 2x^ = (side of sqr)^2.....hence perimeter of sqr = 4*sqrt2*x
circumference of perimeter = pi * (2*x) Thus our eqn becomes => q= pi * (2*x) + 4*sqrt2*x + 4x
on simplification we get x = q / 2(2+2*sqrt(2)+pi) km...OA B Strange, this problem seems to have a different answer when we take the side of the square to be a ( or r) or may be I am doing something wrong! Here, what I am doing  Let the side of the sq be a, then sum of four sides = 4a; Using formula, diag of a sq is sqrt(2)a, we get sum of the paths four paths from castle to moat ( i.e, sum the diagonals) as = 2a sqrt(2) ; Circumference of the circle as 2pie a sqrt(2)/2 Putting it all together we 4a +2asqrt(2) +2pie sqrt(2)/2 = q => 4a +2asqrt(2) +pie asqrt(2) =q => sqrt(2)a (2 sqrt(2)+2 +pie) = q => a = q/sqrt(2) (2 sqrt(2)+2 +pie) Can anybody shed some more light on this..?
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Re: Castle and the path 
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Updated on: 19 Nov 2009, 20:11
rathoreaditya81 wrote: A castle is at the center of the several flat paths which surround it: 4 straight paths that travel from the castle to its circular moat, where they meet up with a perfectly circular path which borders the moat; that circular path circumscribes a square path which has its corners at the ends of the 4 straight paths. If the total length of all of the pathways is q kilometers, then which expression represents distance from the castle to the circular moat?
1) q/4(2+2\sqrt{2}+pie) 2) q/2(2+2\sqrt{2}+pie) 3) q/(2+2\sqrt{2}+pie) 4) 2q/(2+2\sqrt{2}+pie) 5) 4q/2(2+2\sqrt{2}+pie) Let take radius of circular path is R so distance from the castle to the circular moat = R total length of all of the pathways = total length of 4 straight paths that travel from the castle to its circular moat + length of circular path which borders the moat + total length of square path which has its corners at the ends of the 4 straight paths (Diagonal of square is 2R, (904545 right triangle with hypotenuse 2R) so side of square would be \(R\sqrt{2})\) \(q = 4R + 2R*pie + 4 (R \sqrt{2})\) \(q = 2R (2+pie+2 \sqrt{2})\) \(R = q/2 (2+2 \sqrt{2}+ pie)\) Answer B
Originally posted by swatirpr on 18 Nov 2009, 06:36.
Last edited by swatirpr on 19 Nov 2009, 20:11, edited 1 time in total.



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Re: Castle and the path 
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18 Nov 2009, 06:47
swatirpr wrote: rathoreaditya81 wrote: A castle is at the center of the several flat paths which surround it: 4 straight paths that travel from the castle to its circular moat, where they meet up with a perfectly circular path which borders the moat; that circular path circumscribes a square path which has its corners at the ends of the 4 straight paths. If the total length of all of the pathways is q kilometers, then which expression represents distance from the castle to the circular moat?
1) q/4(2+2\sqrt{2}+pie) 2) q/2(2+2\sqrt{2}+pie) 3) q/(2+2\sqrt{2}+pie) 4) 2q/(2+2\sqrt{2}+pie) 5) 4q/2(2+2\sqrt{2}+pie) Let take radius of circular path is R so distance from the castle to the circular moat = R total length of all of the pathways = total length of 4 straight paths that travel from the castle to its circular moat + length of circular path which borders the moat + total length of square path which has its corners at the ends of the 4 straight paths (Diagonal of square is 2R, (904545 right triangle with hypotenuse 2R) so side of square would be R\sqrt{2}) q = 4R + 2R*pie + 4 (R \sqrt{2}) q = 2R (2+pie+2 \sqrt{2}) R = q/2 (2+2 \sqrt{2}+ pie) Answer B Hi Swatipr, have a look at  castlepaths82093.html#p653668
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Re: castle paths
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19 Nov 2009, 20:44
Thanks a lot swatipr.. oblivious to obvious..strange manifestation of alcohol after effect!
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Re: castle paths
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16 Sep 2011, 21:50
can we solve this in more simpler way..say plugging smart nos



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Re: castle paths
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21 May 2014, 17:32
This problem is not that difficult once you understand the question; it just takes too long. Can anyone explain how to do this under 3 minutes?



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Re: castle paths
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21 May 2014, 20:37
TooLong150 wrote: This problem is not that difficult once you understand the question; it just takes too long. Can anyone explain how to do this under 3 minutes? You can actually do it in a minute after you understand that all you need is the radius of the circle given that sum of lengths of all lines and curves here is q. So q is made up of the circle, the inscribed square in it and the 4 radii of the circle. Assume radius of the circle is r. Each side of the square will be \(\sqrt{2}r\) because each side makes an isosceles right triangle with two radii. Each radius is r and sides of isosceles right triangle are in the ratio \(1:1:\sqrt{2}\) q = Circumference of circle + Perimeter of square + 4*radius \(q = 2 \pi r + 4\sqrt{2}r + 4r\) So \(r = q/2( 2 + 2\sqrt{2} + \pi)\) Answer (B)
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Re: A castle is at the center of the several flat paths which su
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02 Nov 2014, 17:04
VeritasPrepKarishma wrote: TooLong150 wrote: This problem is not that difficult once you understand the question; it just takes too long. Can anyone explain how to do this under 3 minutes? You can actually do it in a minute after you understand that all you need is the radius of the circle given that sum of lengths of all lines and curves here is q. So q is made up of the circle, the inscribed square in it and the 4 radii of the circle. Assume radius of the circle is r. Each side of the square will be \(\sqrt{2}r\) because each side makes an isosceles right triangle with two radii. Each radius is r and sides of isosceles right triangle are in the ratio \(1:1:\sqrt{2}\) q = Circumference of circle + Perimeter of square + 4*radius \(q = 2 \pi r + 4\sqrt{2}r + 4r\) So \(r = q/2( 2 + 2\sqrt{2} + \pi)\) Answer (B) Hi Karishma, The part that threw me off was the \(\sqrt{2}r\) I understood that the hypotenuse was \(\sqrt{2}\) but I couldn't connect the R there, and more importantly, I couldn't tell that the problem was asking us to solve for R. Any thoughts on how I would go about this in the future?



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Re: A castle is at the center of the several flat paths which su
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01 Apr 2016, 17:18
wow..the wording is very confusing... total pathways : circumference, the square, and 4 radii. circumference is 2*pi*r square  since we have 4 equilateral triangles, and all right triangles, we can apply 454590 property, and get one side of the square equal to 4*r*sqrt(2) then we have 4r we are then told that: q=2*pi*r + 4*r*sqrt(2) + 4r factor r q=r(2pi+4*sqrt(2)+4) r= q/(2pi+4*sqrt(2)+4)
we can eliminate D and E right away from the above, we can factor out a 2; q/2(pi+2*sqrt(2)+2)
so B.



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Re: A castle is at the center of the several flat paths which su
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