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A circle, with center O, is inscribed in square WXYZ. Point P, as
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Updated on: 13 Jun 2015, 10:35
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A circle, with center O, is inscribed in square WXYZ. Point P, as shown above, is where the circle and the line segment ZX intersect. If PX = 1, which of the following is closest to the value of the circumference of the circle? A. 2π B. 3π C. 4π D. 5π E. 6π Attachment:
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Originally posted by vishalgc on 09 Oct 2008, 21:34.
Last edited by Bunuel on 13 Jun 2015, 10:35, edited 2 times in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.




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Re: A circle, with center O, is inscribed in square WXYZ. Point P, as
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13 Jun 2015, 11:05
petu wrote: Hi, could someone give help me with this one? I just can't understand the explanations! thanks!!! A circle, with center O, is inscribed in square WXYZ. Point P, as shown above, is where the circle and the line segment ZX intersect. If PX = 1, which of the following is closest to the value of the circumference of the circle?A. 2π B. 3π C. 4π D. 5π E. 6π Check image below: Notice that triangle AOX is an isosceles right triangle: AO = AX = radius. We know that hypotenuse of an isosceles right triangle is \(side*\sqrt{2}\), thus = \(OX=side*\sqrt{2}=radius*\sqrt{2}\). On the other hand, OX = OP + PX = radius + PX = radius + 1. Therefore, \(radius + 1=radius*\sqrt{2}\): \(radius=\frac{1}{\sqrt{2}1}=\sqrt{2}+1\approx{1.4+1}=2.4\). The circumference = \(2\pi{r}=2\pi*2.4=4.8\pi\). Answer: D. Hope it's clear. Attachment:
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Re: A circle, with center O, is inscribed in square WXYZ. Point P, as
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09 Oct 2008, 22:32
5pi.
OX = OP + 1 = r + 1 (r is the radius of circle).
Now, OX^2 = r^2 +r^2
Solve for r and then 2pi4 will be approx 5pi.



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Re: A circle, with center O, is inscribed in square WXYZ. Point P, as
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16 Oct 2008, 17:19
scthakur wrote: 5pi.
OX = OP + 1 = r + 1 (r is the radius of circle).
Now, OX^2 = r^2 +r^2
Solve for r and then 2pi4 will be approx 5pi. 5pi Yours is a simple way to solve the problem I added unnecessary steps to it



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Re: A circle, with center O, is inscribed in square WXYZ. Point P, as
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16 Oct 2008, 17:34
scthakur wrote: 5pi.
OX = OP + 1 = r + 1 (r is the radius of circle).
Now, OX^2 = r^2 +r^2
Solve for r and then 2pi4 will be approx 5pi. can you pls explain why you did OX^2=r^2+r^2



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Re: A circle, with center O, is inscribed in square WXYZ. Point P, as
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Updated on: 16 Oct 2008, 19:44
scthakur wrote: 5pi.
OX = OP + 1 = r + 1 (r is the radius of circle).
Now, OX^2 = r^2 +r^2
Solve for r and then 2pi4 will be approx 5pi. Ahhh!! I was not able to see this I was stumped by the question !!
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Originally posted by amitdgr on 16 Oct 2008, 19:43.
Last edited by amitdgr on 16 Oct 2008, 19:44, edited 1 time in total.



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Re: A circle, with center O, is inscribed in square WXYZ. Point P, as
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16 Oct 2008, 19:43
let OP=r the radius, then we know that OP+1=OX
or r+1=OX, lets look at the diagonal of the square ZX, ZX=2OX or 2(r+1)
we also know that ZX=sqrt(2)XY; XY=2R...right..with me???
2r+2=sqrt(2)2r
solve for r, you get r=1/(sqrt(2)1) or 1/0.41
circum. =2pi*r, 2/0.41 is apprx 5...so circum=5pi



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Re: A circle, with center O, is inscribed in square WXYZ. Point P, as
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17 Oct 2008, 01:10
ALD wrote: scthakur wrote: 5pi.
OX = OP + 1 = r + 1 (r is the radius of circle).
Now, OX^2 = r^2 +r^2
Solve for r and then 2pi4 will be approx 5pi. can you pls explain why you did OX^2=r^2+r^2 The distance between midpoint of XY and O is r and distance between midpoint of XY and X is r. That is why, r^2 + r^2 = OX^2.



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Re: A circle, with center O, is inscribed in square WXYZ. Point P, as
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13 Jun 2015, 10:27
Hi, could someone give help me with this one? I just can't understand the explanations! thanks!!!



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Re: A circle, with center O, is inscribed in square WXYZ. Point P, as
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13 Jun 2015, 11:32
THanks Bunuel. Clear as allways. You have few beers waiting for you if you ever come to Barcelona.



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Re: A circle, with center O, is inscribed in square WXYZ. Point P, as
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17 Jan 2017, 14:43
That was an excellent explanation. Thanks Brunel!



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Re: A circle, with center O, is inscribed in square WXYZ. Point P, as
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17 Jan 2017, 14:51
Sorry to ask what might be an obvious questions. From your explanation, I don´t understand the part in bold.
Thanks
Check image below:
Image
Notice that triangle AOX is an isosceles right triangle: AO = AX = radius.
We know that hypotenuse of an isosceles right triangle is side∗2‾√side∗2, thus = OX=side∗2‾√=radius∗2‾√OX=side∗2=radius∗2.
On the other hand, OX = OP + PX = radius + PX = radius + 1.
Therefore, radius+1=radius∗2‾√radius+1=radius∗2:
radius=12‾√−1=2‾√+1≈1.4+1=2.4radius=12−1=2+1≈1.4+1=2.4.
The circumference = 2πr=2π∗2.4=4.8π2πr=2π∗2.4=4.8π.
Answer: D.



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Re: A circle, with center O, is inscribed in square WXYZ. Point P, as
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19 Jan 2017, 07:50
lauramo wrote: Sorry to ask what might be an obvious questions. From your explanation, I don´t understand the part in bold.
Thanks
Check image below:
Image
Notice that triangle AOX is an isosceles right triangle: AO = AX = radius.
We know that hypotenuse of an isosceles right triangle is side∗2‾√side∗2, thus = OX=side∗2‾√=radius∗2‾√OX=side∗2=radius∗2.
On the other hand, OX = OP + PX = radius + PX = radius + 1.
Therefore, radius+1=radius∗2‾√radius+1=radius∗2:
radius=12‾√−1=2‾√+1≈1.4+1=2.4radius=12−1=2+1≈1.4+1=2.4.
The circumference = 2πr=2π∗2.4=4.8π2πr=2π∗2.4=4.8π.
Answer: D. We are solving for radius (you can denote it as x for simplicity) the following equation: \(radius + 1=radius*\sqrt{2}\) to get \(radius=\frac{1}{\sqrt{2}1}=\sqrt{2}+1\approx{1.4+1}=2.4\). \(x + 1=x*\sqrt{2}\) \(1=x*\sqrt{2}x\) \(1=x(\sqrt{2}1)\) \(x=\frac{1}{\sqrt{2}1}\) Multiply both numerator and denominator by \(\sqrt{2}+1\): \(x=\frac{\sqrt{2}+1}{(\sqrt{2}1)(\sqrt{2}+1)}=\sqrt{2}+1\). Hope it's clear.
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