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805+ Level|   Geometry|      
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vishalgc
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A circle, with center O, is inscribed in square WXYZ. Point P, as Updated on: 13 Jun 2015, 10:35
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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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

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A circle, with center O, is inscribed in square WXYZ. Point P, as 13 Jun 2015, 11:05
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petu
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.
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scthakur
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A circle, with center O, is inscribed in square WXYZ. Point P, as 09 Oct 2008, 22:32
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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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A circle, with center O, is inscribed in square WXYZ. Point P, as 16 Oct 2008, 17:19
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scthakur
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 :cool
I added unnecessary steps to it :roll:
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A circle, with center O, is inscribed in square WXYZ. Point P, as 16 Oct 2008, 17:34
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scthakur
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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A circle, with center O, is inscribed in square WXYZ. Point P, as Updated on: 16 Oct 2008, 19:44
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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scthakur
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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A circle, with center O, is inscribed in square WXYZ. Point P, as 16 Oct 2008, 19:43
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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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A circle, with center O, is inscribed in square WXYZ. Point P, as 17 Oct 2008, 01:10
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scthakur
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 mid-point of XY and O is r and distance between mid-point of XY and X is r. That is why, r^2 + r^2 = OX^2.
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A circle, with center O, is inscribed in square WXYZ. Point P, as 13 Jun 2015, 10:27
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Hi, could someone give help me with this one? I just can't understand the explanations! thanks!!!
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A circle, with center O, is inscribed in square WXYZ. Point P, as 13 Jun 2015, 11:32
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THanks Bunuel. Clear as allways. You have few beers waiting for you if you ever come to Barcelona.
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A circle, with center O, is inscribed in square WXYZ. Point P, as 17 Jan 2017, 14:43
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That was an excellent explanation. Thanks Brunel!
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A circle, with center O, is inscribed in square WXYZ. Point P, as 17 Jan 2017, 14:51
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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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A circle, with center O, is inscribed in square WXYZ. Point P, as 19 Jan 2017, 07:50
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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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