Last visit was: 24 Apr 2026, 21:02 It is currently 24 Apr 2026, 21:02
Home
Messages
Tests
Schools
Events
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
605-655 (Medium)|   Geometry|      
avatar
PareshGmat
avatar
Joined: 27 Dec 2012
Last visit: 10 Jul 2016
Posts: 1,531
Own Kudos:
8,274
 [11]
Given Kudos: 193
Status:The Best Or Nothing
Location: India
Concentration: General Management, Technology
WE:Information Technology (Computer Software)
Posts: 1,531
Kudos: 8,274
 [11]
As shown in figure, square ABCD has arc BPD centred at C and arc BQD 10 Sep 2014, 02:21
3
Kudos
Add Kudos
8
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

45% (medium)

Question Stats:

70% (02:24) correct 30% (03:03) wrong based on 179 sessions

History

Date
Time
Result
Not Attempted Yet
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
square.png [ 10 KiB | Viewed 9408 times ]

A: \(16 - 4\pi\)

B: \(8 - 2\pi\)

C: \(8\pi - 16\)

D: \(4\pi - 8\)

E: \(2\pi - 4\)
ShowHide Answer
Official Answer
E
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 24 Apr 2026
Posts: 109,818
Own Kudos:
811,088
 [5]
Given Kudos: 105,873
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,818
Kudos: 811,088
 [5]
As shown in figure, square ABCD has arc BPD centred at C and arc BQD 10 Sep 2014, 04:40
2
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
PareshGmat
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: \(2\pi - 4\)
Show more

First of all, notice that the radii of the circles = the side of the square = 4.

Look at the image below:
Attachment:
Untitled.png
Untitled.png [ 7.93 KiB | Viewed 8140 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 = \(4^2-2*(16 - 4\pi)=8\pi-16\).

The area of the shaded region is 1/4th of the area of the leaf = \(\frac{8\pi-16}{4}=2\pi-4\).

Answer: E.

Hope it's clear.
General Discussion
avatar
nvijayprasanna
avatar
Joined: 17 Apr 2012
Last visit: 10 May 2017
Posts: 12
Own Kudos:
46
 [1]
Given Kudos: 5
Location: United States
WE:Information Technology (Computer Software)
Posts: 12
Kudos: 46
 [1]
As shown in figure, square ABCD has arc BPD centred at C and arc BQD 10 Sep 2014, 06:40
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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.
User avatar
desaichinmay22
User avatar
Joined: 21 Sep 2012
Last visit: 22 May 2016
Posts: 188
Own Kudos:
469
Given Kudos: 31
Location: United States
Concentration: Finance, Economics
Schools: CBS '17
GPA: 4
WE:General Management (Consumer Packaged Goods)
Schools: CBS '17
Posts: 188
Kudos: 469
As shown in figure, square ABCD has arc BPD centred at C and arc BQD 10 Sep 2014, 07:42
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
PareshGmat
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 = \(4^2-2*(16 - 4\pi)=8\pi-16\).

The area of the shaded region is 1/4th of the area of the leaf = \(\frac{8\pi-16}{4}=2\pi-4\).

Answer: E.

Hope it's clear.
Show more

Hi Bunuel,

Can you please explain below part :-
First of all, notice that the radii of the circles = the side of the square = 4.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 24 Apr 2026
Posts: 109,818
Own Kudos:
811,088
 [2]
Given Kudos: 105,873
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,818
Kudos: 811,088
 [2]
As shown in figure, square ABCD has arc BPD centred at C and arc BQD 10 Sep 2014, 07:47
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
desaichinmay22
Bunuel
PareshGmat
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 = \(4^2-2*(16 - 4\pi)=8\pi-16\).

The area of the shaded region is 1/4th of the area of the leaf = \(\frac{8\pi-16}{4}=2\pi-4\).

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.
Show more

We are told that arc BPD centred at C:
Attachment:
Untitled.png
Untitled.png [ 10.44 KiB | Viewed 8014 times ]
User avatar
desaichinmay22
User avatar
Joined: 21 Sep 2012
Last visit: 22 May 2016
Posts: 188
Own Kudos:
469
Given Kudos: 31
Location: United States
Concentration: Finance, Economics
Schools: CBS '17
GPA: 4
WE:General Management (Consumer Packaged Goods)
Schools: CBS '17
Posts: 188
Kudos: 469
As shown in figure, square ABCD has arc BPD centred at C and arc BQD 10 Sep 2014, 08:09
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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.
avatar
PareshGmat
avatar
Joined: 27 Dec 2012
Last visit: 10 Jul 2016
Posts: 1,531
Own Kudos:
8,274
Given Kudos: 193
Status:The Best Or Nothing
Location: India
Concentration: General Management, Technology
WE:Information Technology (Computer Software)
Posts: 1,531
Kudos: 8,274
As shown in figure, square ABCD has arc BPD centred at C and arc BQD 25 Sep 2014, 21:18
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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
square.png [ 10.81 KiB | Viewed 7761 times ]
Consider any Quarter circle

\(Its Area = \frac{\pi4^2}{4} = 4x + 8 - 2x\)

\(x = 2\pi - 4\)

Answer = E
avatar
Ashishsteag
avatar
Joined: 28 Dec 2015
Last visit: 17 Nov 2016
Posts: 26
Own Kudos:
13
Given Kudos: 62
Posts: 26
Kudos: 13
As shown in figure, square ABCD has arc BPD centred at C and arc BQD 08 Jul 2016, 03:04
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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)=16-4pie

Similarly,Area under the second quarter arc=4 pie.

Subtract it from the Area of square,you will get the remaining Area(ABPDA)=16-4pie

Add both of these areas=32-8pie.

Now that you get the areas of these two portions,subtract it from the Area of square=16-(32-8pie)=8pie-16

So,area of just 1/4th section=1/4*8pie-16=2pie-4
avatar
abhi_abhu
avatar
Joined: 15 Jul 2013
Last visit: 14 Jun 2020
Posts: 2
Given Kudos: 7
Posts: 2
Kudos: 0
As shown in figure, square ABCD has arc BPD centred at C and arc BQD 08 Jul 2016, 04:10
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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
User avatar
Divyadisha
User avatar
Current Student
Joined: 18 Oct 2014
Last visit: 01 Jun 2018
Posts: 660
Own Kudos:
1,958
Given Kudos: 69
Location: United States
GMAT 1: 660 Q49 V31
GPA: 3.98
GMAT 1: 660 Q49 V31
Posts: 660
Kudos: 1,958
As shown in figure, square ABCD has arc BPD centred at C and arc BQD 08 Jul 2016, 05:18
Kudos
Add Kudos
Bookmarks
Bookmark this Post
PareshGmat
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: \(2\pi - 4\)
Show more

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= 4pi-8/2= 2pi-4

E is the answer
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,976
Own Kudos:
1,117
Posts: 38,976
Kudos: 1,117
As shown in figure, square ABCD has arc BPD centred at C and arc BQD 07 Apr 2018, 08:28
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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.
Moderators:
Bunuel
Math Expert
109818 posts
Krunaal
Tuck School Moderator
853 posts