The above figure shows a circle with center O.

Question

GMATBusters’ Quant Quiz Question -2

For past quiz questions, click here

The above figure shows a circle with center O. ACEO is a Square. what is the area of the circle?
1) The perimeter of the shaded region is (pi+2.828)
2) Area of the shaded region is (pi-2)

Show Spoiler

Attachment:

Capture.JPG [ 18.37 KiB | Viewed 4491 times ]

Show Answer

Official Answer

D

GMATBusters · Answer

The Solution of the Question shall be as follows:
 https://www.youtube.com/watch?v=tYCJl6dz57s

ashwini2k6jha · Answer

area =?
state1:sufficient
perimeter=pi+2.828
x+r-x+x+r-x=pi+2.828
r=pi/2+2.828
area=pi x r^2 sufficient
state:2
area shaded=pi-2
2x 1/2x(r-x)xx=pi-2
(r-x)x=pi-2 not sufficient
Answer A

Raxit85 · Answer

The above figure shows a circle with center O. what is the area of the circle?
1) The perimeter of the shaded region is (pi+2.828) -
Central angle/360 = perimeter of segment/perimeter of circle
1/4 = pi+2.828/2 * pi * r
2*pi*r = 4pi+2*2.828.
r can be found with calculation. So area of the circle can also found.
Sufficient.

2) Area of the shaded region is (pi-2)
Not sufficient.

Ans A.

uc26 · Answer

In the given figure, we can see a square AOEC is inscribed in a quarter of a circle with center O. If we can find the radius of the circle that would be sufficient to find the area of the circle=  pi*r^2 . Also, since angle BOD is 90 degrees, it forms a quarter of the circle.

1) Perimeter of the shaded region would mean the circumference of the circle from B to D which constitutes for a quarter of the total circumference.  1/4*2*pi*r=pi+2.828 . Since we know pi is approx 3.14 or 22/7, we can find the radius of the circle. Sufficient.

(2) Area of the shaded region= Area of the quarter of the circle- Area of the square AOEC. We can draw a diagonal (d) OC for the square AOEC joining O and C. Notice that OC is also the radius of the circle. Let the side of square= a. We know that diagonal of a square (d)= \sqrt{2}*a. So  a^2= r^2/2 . Area of a square is  side^2= a^2= r^2/2 . Area of shaded region=  pi-2= 1/4* pi* r^2- r^2/2 . Since we we know that pi is approx 3.14 or 22/7, we can find the radius of the circle. Sufficient.

Answer: D.

Tommy6e · Answer

Answer is D

Statement 1:
Angle BOD is 90, which is 360/4 --> the perimeter of BCD represents half of the perimeter of the circle --> The perimeter of the circle is 4 times pi+2.828 --> 4pi + 4*2.828 = 2pi*Radius --> we can find the Radius and thus the area of the circle (Area = Radius * pie^2)

Statement 2:
AOEC is a square, thus triangle OCE is a right isosceles triangle. If OE=EC=X, then 2X^2 = R^2 (see screenshot.)
Then the area of square AOCE is x^2=R^2/2 --> The area of CMNK is 4 * R^2/2 = 2R^2.
The area of the shaded yellow region is 4*(pie-2), which is 4 times the original shaded region.
The total area of the circle is 2R^2 (square MNKC) plus 4*(pie-2) (yellow shaded region.) This can be equal to pie*Radius^2. Based on this formula we can find the Radius and thus the area of the circle.

Archit3110 · Answer

The above figure shows a circle with center O. what is the area of the circle?
1) The perimeter of the shaded region is (pi+2.828)
2) Area of the shaded region is (pi-2)
let side of square (OE) = s and radius of OD = r

#1
The perimeter of the shaded region is (pi+2.828)
the said perimeter can be re written as pi + 3 
which would be for circle side 1/4
1/4 * 2pi*r - ( perimeter of square AOCE) = pi+3
pi*r-12/8 = s ---(1)
digonal of square = r√2=s
determine r and area can be determined
sufficient
#2
Area of the shaded region is (pi-2)
1/4 * pi *r^2 - (s^2) = pi-2
use r√2=s
sufficient 
determine r then area can be determined
sufficient
OPTION D

Kinshook · Answer

Given: The above figure shows a circle with center O.

Asked: what is the area of the circle?

Area of the circle =  \pi r^2

1) The perimeter of the shaded region is  (\pi+2.828) 
Perimeter of shaded region =  2r + \pi r/2 = \pi+2.828 
Radius of the circle r is known
Area of the circle is known
SUFFICIENT

2) Area of the shaded region is  \pi-2 
Let us assume side of the square = x
Radius of the circle  r = x\sqrt{2} 
Area of shaded region =  \pi r^2 /4 - x^2 = (\pi/4 - 1/2)r^2 = \pi - 2 
r^2 = 4; r = 2
Area of the circle =  \pi r^2 = 4 \pi  
SUFFICIENT

IMO D

Vz_WIbnJBYI

1BlackDiamond1 · Answer

So the question is to find the area of a circle which is pi*R^2. So the hidden question is to find the radius of the circle.
Statement 1:
1) The perimeter of the shaded region is (pi+2.828)

let Side of square is 'a'. Then by Pythagorean theorem, we get
2a^2=R^2
therefore a = R/root2

The perimeter of the shaded region = (1/4 * 2 * pi * R)+ 2 *a + 2(R-a)
since a= R/root2
(pi+2.828) = (1/4 * 2 * pi * R)+ 2 * R/root2 + 2(R-R/root2)

This is equation with one variable. No need to solve further.
Hence Statment 1 is sufficient.

2) Area of the shaded region is (pi-2)

Area of the shaded region = 1/4*pi*R^2 -(side of the square)^2
let Side of square is 'a'. Then by Pythagorean theorem, we get
2a^2=R^2
therefore a = R/root2
substituting this value in the above equation we get
(pi-2) =(1/4* pi* R^2) - (R/ root2)

This is an equation with one variable hence we can find the value of R. No need to solve further since we can find R.

Thus statement 2 is sufficient.

This answer is D.

SR71Blackbird · Answer

The answer to this question is D

Statement 1: perimeter of shaded region is pi + 2.828. Perimeter is made up of two sides of square + quarter of circumference + two (radius - side of square) extra lengths.

Let the side of the square be x, then radius is x*2^(1/2), extra length is x*2^(1/2)-x 
x + x + x*2^(1/2)-x + x*2^(1/2)-x + (2 *pi* x*2^(1/2))/4 = pi + 2.828. Solving this will give us the value of the x, then radius and area

Statement 2: Area of the shaded region = pi-2 
Shaded region = Quarter circle area - square 
Shaded region = (pi * r^2)/4 - x^2, radius = x*2^(1/2) = pi - 2, solve for x we get side of square, radius and area of complete circle

kapilagarwal · Answer

Answer : D Each Alone sufficient. From both we can both Radius

Posted from my mobile device

zs2 · Answer

The above figure shows a circle with center O. ACEO is a Square. what is the area of the circle?
1) The perimeter of the shaded region is (pi+2.828)
2) Area of the shaded region is (pi-2)

Let side of square = x 
and radius of circle = r

we need to find r to find area of circle.

r is also the diagonal of square so r = x sq root 2

Statement 1- 
x+x+r-x+r-x+(2.pi.r)/4 = pi+2.828
we can find r from this as we know relation between x and r , hence 1 sufficient

Statement -2 
area of shaded region is pi-2

that is: area of quarter circle - area of square = pi-2
pi.r^2/4 - x^2 = pi-2 
we can again find r here as we know the relation between x and r . Hence 2 is also sufficient.

Hence D is answer.

lacktutor · Answer

The above figure shows a circle with center O. ACEO is a Square. what is the area of the circle?

Let's say that the side of a square and the radius of a circle are x and R, respectively.
--> Radius = ?
(Statement1): The perimeter of the shaded region is (π+2.828)
-->  (R-x)+ x+x +(R-x) + \frac{2πR}{4}= π+2.828 
 2R (1+π/4)= π+ 2.828 
 R= \frac{(π+ 2.828)}{(1+π/4)} 
Sufficient

(Statement2): Area of the shaded region is (π-2)
 2x^{2}= R^{2} 
-->  x^{2}= \frac{R^{2}}{2}

\frac{πR^{2}}{4} - x^{2}= π-2 
 \frac{πR^{2}}{4} - \frac{R^{2}}{2} =π-2 
 \frac{R^{2} (π-2)}{4} = π-2 
-->  R=2 
Sufficient

Answer (D)

jrk23 · Answer

i think answer is "D" as we can make two equation from both of the.

Just need to use that here side of square is r/2^1/2 (here r is the radius of the circle)

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.

The above figure shows a circle with center O.

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 Data Sufficiency (DS) Questions

The above figure shows a circle with center O.

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