In the figure above, a square is inscribed in a circle. If the area of

Question

https://gmatclub.com/forum/download/file.php?id=30439 In the figure above, a square is inscribed in a circle. If the area of the shaded region is 4π - 8, what is the value of the radius? (A) 1 (B) 2 (C) 2π (D) 2√2 (E) 4 circle_inscribed_square.png

FacelessMan · Answer

A general equation for the area of the shaded region = πr^2 - 2r^2, where r = radius of the circle.
equating this to the given value => πr^2 - 2r^2 = 4π - 8, Thus r = 2.

Hence B.

CounterSniper · Answer

substitute--

if r = 2
diagonal of square = Xroot2 where X is the side 
X\sqrt{2}=4 --> X= 4/\sqrt{2} , area = 16/2=8

now area of circle = 4π

which satisfies the condition

area of the shaded region is 4π - 8

rohit8865 · Answer

Let sides of square=S and radius of circle=r
as diagonal is making relation with sides of square S and radius of r 
we will try to find relation between them 
Diagonal =sq.root2*S
also diagonal=2r
sq.root2S=2r
S=sq. root 2 *r
area of square= 2r^2
area of circle=πr^2
area of circle-area of square=4π-8(given)
πr^2-2*r^2=4π-8
r^2=4
r=2
Ans B

sadikabid27 · Answer

JeffTargetTestPrep   Bunuel  please share your approach here. Thanks.

Bunuel · Answer

https://gmatclub.com/forum/download/file.php?id=30439 
 In the figure above, a square is inscribed in a circle. If the area of the shaded region is 4π - 8, what is the value of the radius?

(A) 1
(B) 2
(C) 2π
(D) 2√2
(E) 4

Let d be the length of a diameter of the circle, which is also the diagonal of the square.

The area of the circle is  \pi*(\frac{diameter}{2})^2=\pi*\frac{d^2}{4} .

The area of the square is  \frac{diagonal^2}{2}=\frac{d^2}{2} .

The area of the shaded region is  \pi*\frac{d^2}{4}-\frac{d^2}{2}=4\pi-8 ;

d^2(\frac{\pi - 2}{4})=4(\pi-2) ;

d^2=16 ;

d = 4 ;

radius=\frac{d}{2}=2 .

Answer: B.

khan0210 · Answer

Fair to say, I perhaps arrived at the solution by fluke but:

Pi.r^2 - s^2
given pi(4)-8
:. pi(2^2)-(\sqrt{8})^2

so radius is 2.

ScottTargetTestPrep · Answer

The diameter of the circle is equal to the diagonal of the square.Let’s let 2r = diagonal = diameter and r = the radius of the circle.

Since diagonal = side√2, we have:

2r = s√2

2r/√2 = s

So the area of the square = s^2 = (4r^2)/2 = 2r^2

The area of the shaded region is found by subtraction: area of circle - area of square = area of shaded region. So finally, we have:

πr^2 - 2r^2 = 4π - 8

r^2(π - 2) = 4(π - 2)

r^2 = 4

r = 2

Answer: B

CoachNikhilesh · Answer

Here,
Let the side be s
and let the radius be r

The diameter will be,
 2r = \sqrt{s^2+s^2} 
 2r = s\sqrt{2} 
 s = \sqrt{2} r

Area of the shaded region,
 \pi r^2 - 2 r^2 = 4 \pi - 8

Substituting values from options

(A) r = 1, gives  \pi - 2 
(B) r = 2, gives  4\pi - 8

Therefore,  the answer is (B)

bobnil · Answer

Area of shaded region = (area of the circle) - (area of the square)
For inscribed square, the radius of circle = diagonal of the square . 
So,  4r^2= a^2+a^2  ( pythergorian theorem)
 a^2 = 2r^2   = ( area of square)
-->  (\pi * r^2)-2r^2 = 4\pi – 8  
--> r=2
ans = B

imshizzy · Answer

I may have had a bit of luck on my side but when I first looked at the problem I noticed the shaded region was full of irregular shapes, therefore I reasoned that the easiest method to get the area of the shaded region would be subtracting the area of the square from the area of the circle. Looking at the equation that was given (4π-8) that looked spot on...a formula that looks like the area of a circle subtracting an integer which could be assumed to be the area of a square.

i set up the following equation 4π=πr^2. Cancel out the πs and you are left with 4=r^2 which gives the correct answer of two.

If you work backwards using the area of the square, this method also checks out.

i set up the following equation 4π=πr^2. Cancel out the πs and you are left with 4=r^2 which gives the correct answer of two.

If you work backwards using the area of the square, this method also checks out.

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

In the figure above, a square is inscribed in a circle. If the area of

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

In the figure above, a square is inscribed in a circle. If the area of

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards