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A square is inscribed within a circle, as shown above. If the total ar

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A square is inscribed within a circle, as shown above. If the total ar  [#permalink]

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New post Updated on: 31 May 2015, 06:00
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A square is inscribed within a circle, as shown above. If the total area of the shaded regions is 2, what is the area of the circle?

A. 2pi/(pi-2)
B. 2pi/p-1)
C. pi/(pi-2)
D. pi/(pi-4)
E. 2pi/(pi-4)

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Originally posted by reto on 30 May 2015, 06:22.
Last edited by Bunuel on 31 May 2015, 06:00, edited 1 time in total.
Edited the question.
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Re: A square is inscribed within a circle, as shown above. If the total ar  [#permalink]

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New post 30 May 2015, 07:20
HI. I am getting the answer 4PI/PI-2.

Suppose side of the square=a. then diagnal= \sqrt{2} X a.

Divide the diagnoal by 2= You will be get= \sqrt{2} Xa /2= a/\sqrt{2}

i am getting the answer 4PI/PI-2.


reto wrote:
Attachment:
T8916.png

A square is inscribed within a circle, as shown above. If the total area of the shaded regions is 2, what is the area of the circle?

A. 2pi/pi-2
B. 2pi/p-1
C. pi/pi-2
D. pi/pi-4
E. 2pi/pi-4

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Re: A square is inscribed within a circle, as shown above. If the total ar  [#permalink]

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New post 30 May 2015, 09:16
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Lets assume side of the square = \(a\), diagonal is \(a*\sqrt{2}\) so radius is half of it - \(\frac{a}{2}*\sqrt{2}\)
\(2 = \pi*\frac{a^2}{2} - a^2\) (1)
What we need to find is: S = \(\pi*\frac{a^2}{2}\)
From the (1): \(a^2 = 2/(\pi/2 - 1)\)
put it in our resulting equation for the answr:
\(S = \pi/2*2/(\pi/2 - 1)= 2*\pi/(\pi - 2)\) which corresponds to A
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Re: A square is inscribed within a circle, as shown above. If the total ar  [#permalink]

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New post 31 May 2015, 01:55
Another way to solve such question with shaded regions is to ballpark the answer choices. In the GMAT questions with shaded areas are there because these areas has no names.

Start with a snapshot ballpark. If the shaded region is 2, estimate how big the whole circle must be. In this example it's about 6 from guessing.

With this you can already POE all answer choices but A.

Generally for shaded regions in the GMAT:

Shaded regions - Subtraction with pi
• In the GMAT, an area is shaded when there is no geometric term to describe it directly.
• In such cases that area can be calculated as a subtraction of one shape from another.
• In the case there is a circular rim and a straight edge - The GMAT uses a subtraction involving a circle and a non-circular shape.
• Since does not cancel out, whenever an exact result is required you can:
o POE answers without
o POE answers without subtraction
o And of course- as always, POE answers out of the ballpark
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Re: A square is inscribed within a circle, as shown above. If the total ar  [#permalink]

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New post 31 May 2015, 02:49
Hi shriramvelamuri,

Did the above explanations clear this up for you? I was trying to see where your approach broke down, but I don't see how you got from diagonal = a/2rt.2 to your answer.
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New post 31 May 2015, 05:59
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New post 26 Jul 2015, 19:16
Hey Bunuel,

Came across this question today and could not figure it out. Can you break it down for us? Appreciated.
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Re: A square is inscribed within a circle, as shown above. If the total ar  [#permalink]

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New post 26 Jul 2015, 19:39
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sajjad254 wrote:
Hey Bunuel,

Came across this question today and could not figure it out. Can you break it down for us? Appreciated.


Let r be the radius of the circle and 'a' be the side of the square.

As shown in the attached figure, in the right triangle drawn:

\(r^2 = (a/2)^2 + (a/2)^2\) ---> \(a^2 = 2r^2\)

From the given figure,

Shaded area = area of the circle - area of the square ---> \(\pi * r^2 - a^2 = 2\) ----> \(\pi * r^2 - 2r^2 = 2\) ----> \(r^2 = \frac{2}{\pi -2}\)

Thus, area of the circle = \(\pi*r^2 = \frac{2\pi}{\pi -2}\) . A is the correct answer.
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Re: A square is inscribed within a circle, as shown above. If the total ar  [#permalink]

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New post 27 Jul 2015, 07:39
Engr2012 wrote:
sajjad254 wrote:
Hey Bunuel,

Came across this question today and could not figure it out. Can you break it down for us? Appreciated.


Let r be the radius of the circle and 'a' be the side of the square.

As shown in the attached figure, in the right triangle drawn:

\(r^2 = (a/2)^2 + (a/2)^2\) ---> \(a^2 = 2r^2\)

From the given figure,

Shaded area = area of the circle - area of the square ---> \(\pi * r^2 - a^2 = 2\) ----> \(\pi * r^2 - 2r^2 = 2\) ----> \(r^2 = \frac{2}{\pi -2}\)

Thus, area of the circle = \(\pi*r^2 = \frac{2\pi}{\pi -2}\) . A is the correct answer.



Thank you! Super clear now.
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A square is inscribed within a circle, as shown above. If the total ar  [#permalink]

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New post 21 Sep 2015, 03:01
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Area of the shaded region is equal to: \(∏r^2 (circle)-a^2 (square)\)
Since we know that 2r is diagonal of the square, we can get the side of the square by using Pythagorem Theorem:
\(a^2+a^2=〖(2r)〗^2\)
\(2a^2 = 4r^2\)
\(a^2 = 2r^2\)
\(a=√2rr\)
\(a=r*√2\)

The shaded area equals to 2. So:

\(∏r^2-a^2=2\)
\(∏r^2- (r*√2)^2=2\)
\(∏r^2-r^2*2=2\)
\(r^2*(∏-2)=2\)
\(r^2=\frac{2}{(∏-2)}\)

So the area of the circle is equal to \(r^2*∏\)
That gives us area of the circle = \(\frac{2}{(∏-2)} *∏\)

So area of the circle is \(\frac{(2∏)}{((∏-2))}\)
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Re: A square is inscribed within a circle, as shown above. If the total ar  [#permalink]

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New post 03 Dec 2016, 04:24
If r is the radius of a circle and a square is inscribed in it, the area of the square is 2r^2. Why? Because r is the radius and there can be 4 triangles formed inside such a square when diagonals (diameters) cross each other. Each triangle has an area = 1/2 r*r = > 1/2r^2. So the square's area = 2r^2.

area of circle - area of square = 2
if we re-arrange r^2 to one side of the equation, we will get choice A.
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Re: A square is inscribed within a circle, as shown above. If the total ar  [#permalink]

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New post 16 Jul 2018, 08:40
reto wrote:
Attachment:
T8916.png

A square is inscribed within a circle, as shown above. If the total area of the shaded regions is 2, what is the area of the circle?

A. 2pi/(pi-2)
B. 2pi/p-1)
C. pi/(pi-2)
D. pi/(pi-4)
E. 2pi/(pi-4)


A general equation for the area of the shaded region
= πr^2 - 2r^2, where r = radius of the circle.

So, πr^2 - 2r^2=2
r^2=2/pi-2
πr^2=2pi/(pi-2)
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A square is inscribed within a circle, as shown above. If the total ar  [#permalink]

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New post 25 Jul 2018, 12:00
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reto wrote:
Attachment:
T8916.png

A square is inscribed within a circle, as shown above. If the total area of the shaded regions is 2, what is the area of the circle?

A. 2pi/(pi-2)
B. 2pi/p-1)
C. pi/(pi-2)
D. pi/(pi-4)
E. 2pi/(pi-4)



can also be calculated as

The area of the shaded region/4 = area of sector - area of triangle

since we have a square and each of the shaded region has same area
we pick one of the shaded region
area of one such shaded region = 2/4
also,
let a be the side of the square
the diagonal therefore will be a*root2 = diameter of the circle
radius will be a*root2/2
so
2/4=pi*r^2*90/360 - 1/2(a*root2/2)*(a*root2/2)
2/4=p1*2a^2/16 - a^2/4
this gives
a=4/2pi-1
now area of the circle = pi*r^2
and r = a*root2/2
substitute for r
we get 2pi/2pi-1
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A square is inscribed within a circle, as shown above. If the total ar   [#permalink] 25 Jul 2018, 12:00
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