May 25 07:00 AM PDT  09:00 AM PDT Attend this webinar and master GMAT SC in 10 days by learning how meaning and logic can help you tackle 700+ level SC questions with ease. May 24 10:00 PM PDT  11:00 PM PDT Join a FREE 1day workshop and learn how to ace the GMAT while keeping your fulltime job. Limited for the first 99 registrants. May 27 01:00 AM PDT  11:59 PM PDT All GMAT Club Tests are free and open on May 27th for Memorial Day! May 27 10:00 PM PDT  11:00 PM PDT Special savings are here for Magoosh GMAT Prep! Even better  save 20% on the plan of your choice, now through midnight on Tuesday, 5/27 May 30 10:00 PM PDT  11:00 PM PDT Application deadlines are just around the corner, so now’s the time to start studying for the GMAT! Start today and save 25% on your GMAT prep. Valid until May 30th. Jun 01 07:00 AM PDT  09:00 AM PDT Learn reading strategies that can help even nonvoracious reader to master GMAT RC
Author 
Message 
TAGS:

Hide Tags

SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1812
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)

As shown in figure, square ABCD has arc BPD centred at C and arc BQD
[#permalink]
Show Tags
10 Sep 2014, 02:21
Question Stats:
69% (02:26) correct 31% (02:59) wrong based on 165 sessions
HideShow timer Statistics
As shown in figure, square ABCD has arc BPD centred at C and arc BQD centred at A. If AB = 4, the area of the shaded region is Attachment:
square.png [ 10 KiB  Viewed 5478 times ]
A: \(16  4\pi\) B: \(8  2\pi\) C: \(8\pi  16\) D: \(4\pi  8\) E: \(2\pi  4\)
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Kindly press "+1 Kudos" to appreciate




Math Expert
Joined: 02 Sep 2009
Posts: 55271

Re: As shown in figure, square ABCD has arc BPD centred at C and arc BQD
[#permalink]
Show Tags
10 Sep 2014, 04:40
PareshGmat wrote: As shown in figure, square ABCD has arc BPD centred at C and arc BQD centred at A. If AB = 4, the area of the shaded region is A: \(16  4\pi\) B: \(8  2\pi\) C: \(8\pi  16\) D: \(4\pi  8\) E: \(2\pi  4\) First of all, notice that the radii of the circles = the side of the square = 4. Look at the image below: Attachment:
Untitled.png [ 7.93 KiB  Viewed 4642 times ]
When we subtract the area of DBC, which is 1/4th of the circle, we get the area of the red portion > red = \(4^2  \frac{\pi{r^2}}{4}=16  4\pi\). If we subtract twice of that from the area of the square, we get the area of the leaf shaped figure in the centre > the area of the leaf = \(4^22*(16  4\pi)=8\pi16\). The area of the shaded region is 1/4th of the area of the leaf = \(\frac{8\pi16}{4}=2\pi4\). Answer: E. Hope it's clear.
_________________




Intern
Joined: 17 Apr 2012
Posts: 14
Location: United States
WE: Information Technology (Computer Software)

Re: As shown in figure, square ABCD has arc BPD centred at C and arc BQD
[#permalink]
Show Tags
10 Sep 2014, 06:40
Thanks Bunuel,
Another method of solving. Consider the arc BPD with center at C. It is 1/4th of circle with center at C and radius of BC that is side of the square  4. So the area of the BPDC is 1/4 * pi * 4* 4 = 4*pi.
in the arc, BCD is right angle triangle at C, base and height will be side of the square. Hence area of triangle BCD will be 1/2 * 4 * 4 = 8.
difference between these two area will give Arc BPD with base of BD = 4pi  8. half of it will be 2pi  4. which will be area of the shaded region.



Manager
Joined: 21 Sep 2012
Posts: 213
Location: United States
Concentration: Finance, Economics
GPA: 4
WE: General Management (Consumer Products)

Re: As shown in figure, square ABCD has arc BPD centred at C and arc BQD
[#permalink]
Show Tags
10 Sep 2014, 07:42
Bunuel wrote: PareshGmat wrote: As shown in figure, square ABCD has arc BPD centred at C and arc BQD centred at A. If AB = 4, the area of the shaded region is A: \(16  4\pi\) B: \(8  2\pi\) C: \(8\pi  16\) D: \(4\pi  8\) E: \(2\pi  4\) First of all, notice that the radii of the circles = the side of the square = 4. Look at the image below: Attachment: Untitled.png When we subtract the area of DBC, which is 1/4th of the circle, we get the area of the red portion > red = \(4^2  \frac{\pi{r^2}}{4}=16  4\pi\). If we subtract twice of that from the area of the square, we get the area of the leaf shaped figure in the centre > the area of the leaf = \(4^22*(16  4\pi)=8\pi16\). The area of the shaded region is 1/4th of the area of the leaf = \(\frac{8\pi16}{4}=2\pi4\). Answer: E. Hope it's clear. Hi Bunuel, Can you please explain below part : First of all, notice that the radii of the circles = the side of the square = 4.



Math Expert
Joined: 02 Sep 2009
Posts: 55271

Re: As shown in figure, square ABCD has arc BPD centred at C and arc BQD
[#permalink]
Show Tags
10 Sep 2014, 07:47
desaichinmay22 wrote: Bunuel wrote: PareshGmat wrote: As shown in figure, square ABCD has arc BPD centred at C and arc BQD centred at A. If AB = 4, the area of the shaded region is A: \(16  4\pi\) B: \(8  2\pi\) C: \(8\pi  16\) D: \(4\pi  8\) E: \(2\pi  4\) First of all, notice that the radii of the circles = the side of the square = 4. Look at the image below: Attachment: The attachment Untitled.png is no longer available When we subtract the area of DBC, which is 1/4th of the circle, we get the area of the red portion > red = \(4^2  \frac{\pi{r^2}}{4}=16  4\pi\). If we subtract twice of that from the area of the square, we get the area of the leaf shaped figure in the centre > the area of the leaf = \(4^22*(16  4\pi)=8\pi16\). The area of the shaded region is 1/4th of the area of the leaf = \(\frac{8\pi16}{4}=2\pi4\). Answer: E. Hope it's clear. Hi Bunuel, Can you please explain below part : First of all, notice that the radii of the circles = the side of the square = 4. We are told that arc BPD centred at C: Attachment:
Untitled.png [ 10.44 KiB  Viewed 4594 times ]
_________________



Manager
Joined: 21 Sep 2012
Posts: 213
Location: United States
Concentration: Finance, Economics
GPA: 4
WE: General Management (Consumer Products)

Re: As shown in figure, square ABCD has arc BPD centred at C and arc BQD
[#permalink]
Show Tags
10 Sep 2014, 08:09
Hi Bunuel, Can you please explain below part : First of all, notice that the radii of the circles = the side of the square = 4.[/quote] We are told that arc BPD centred at C: Attachment: Untitled.png [/quote] As usual grateful to you for the help. Thanks a ton.



SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1812
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)

Re: As shown in figure, square ABCD has arc BPD centred at C and arc BQD
[#permalink]
Show Tags
25 Sep 2014, 21:18
Refer diagram below: Let the area of the required shaded region = x Area of all other regions shaded would be as shown in the diagram as area of the full square = 16 Attachment:
square.png [ 10.81 KiB  Viewed 4349 times ]
Consider any Quarter circle \(Its Area = \frac{\pi4^2}{4} = 4x + 8  2x\) \(x = 2\pi  4\) Answer = E
_________________
Kindly press "+1 Kudos" to appreciate



Intern
Joined: 28 Dec 2015
Posts: 39

Re: As shown in figure, square ABCD has arc BPD centred at C and arc BQD
[#permalink]
Show Tags
08 Jul 2016, 03:04
I solved it this way:
Area of square=16
Area under one quarter arc=90/360*pie*4^2=4pie.
Subtract it from the Area of square,we will get Area of remaining portion(DCBQD)=164pie
Similarly,Area under the second quarter arc=4 pie.
Subtract it from the Area of square,you will get the remaining Area(ABPDA)=164pie
Add both of these areas=328pie.
Now that you get the areas of these two portions,subtract it from the Area of square=16(328pie)=8pie16
So,area of just 1/4th section=1/4*8pie16=2pie4



Intern
Joined: 15 Jul 2013
Posts: 2

As shown in figure, square ABCD has arc BPD centred at C and arc BQD
[#permalink]
Show Tags
08 Jul 2016, 04:10
let the point of intersection of diagonals be O. Now the area of the shaded region is (area of sector BPC  area of triangle BOC)
area of sector = [pi/4][/2*p] * pi * r2 = 2pi area of triangle BOC = 16/4 = 4
result : 2pi  4 Option E



Current Student
Joined: 18 Oct 2014
Posts: 834
Location: United States
GPA: 3.98

Re: As shown in figure, square ABCD has arc BPD centred at C and arc BQD
[#permalink]
Show Tags
08 Jul 2016, 05:18
PareshGmat wrote: As shown in figure, square ABCD has arc BPD centred at C and arc BQD centred at A. If AB = 4, the area of the shaded region is Attachment: square.png A: \(16  4\pi\) B: \(8  2\pi\) C: \(8\pi  16\) D: \(4\pi  8\) E: \(2\pi  4\) That is how I solved it. Area of Square= 4*4 = 16 Area of half square= 8 Area of one arc = 90/360 * pi *4*4= 4pi area of shaded region= 4pi8/2= 2pi4 E is the answer
_________________
I welcome critical analysis of my post!! That will help me reach 700+



NonHuman User
Joined: 09 Sep 2013
Posts: 11012

Re: As shown in figure, square ABCD has arc BPD centred at C and arc BQD
[#permalink]
Show Tags
07 Apr 2018, 08:28
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________




Re: As shown in figure, square ABCD has arc BPD centred at C and arc BQD
[#permalink]
07 Apr 2018, 08:28






