A circle, with center O, is inscribed in square WXYZ. Point P, as

Question

https://gmatclub.com/forum/download/file.php?id=26945 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 Picture2-1%5B1%5D.png

Bunuel · Accepted Answer

https://gmatclub.com/forum/download/file.php?id=26945 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: https://gmatclub.com/forum/download/file.php?id=26946 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. Untitled.png

scthakur · Answer

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.

LiveStronger · Answer

5pi Yours is a simple way to solve the problem :cool

I added unnecessary steps to it :roll:

ALD · Answer

can you pls explain why you did OX^2=r^2+r^2

amitdgr · Answer

Ahhh!! I was not able to see this

I was stumped by the question !!

FN · Answer

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

scthakur · Answer

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.

petu · Answer

Hi, could someone give help me with this one? I just can't understand the explanations! thanks!!!

petu · Answer

THanks Bunuel. Clear as allways. You have few beers waiting for you if you ever come to Barcelona.

lauramo · Answer

That was an excellent explanation. Thanks Brunel!

lauramo · Answer

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.

Bunuel · Answer

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.

bumpbot · Answer

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A circle, with center O, is inscribed in square WXYZ. Point P, as

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