Jul 21 07:00 AM PDT  09:00 AM PDT Attend this webinar to learn a structured approach to solve 700+ Number Properties question in less than 2 minutes Jul 20 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. Jul 26 08:00 AM PDT  09:00 AM PDT The Competition Continues  Game of Timers is a teambased competition based on solving GMAT questions to win epic prizes! Starting July 1st, compete to win prep materials while studying for GMAT! Registration is Open! Ends July 26th Jul 27 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: 1787
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)

Square ABCD has arc AC centred at B and arc BD centred at C
[#permalink]
Show Tags
10 Sep 2014, 22:22
Question Stats:
62% (02:41) correct 38% (02:51) wrong based on 112 sessions
HideShow timer Statistics
As shown in figure, Square ABCD has arc AC centred at B and arc BD centred at C. If AB = 4, area of shaded region is A: \(\frac{16\pi}{3}  4\sqrt{3}\) B: \(4\sqrt{3}  \frac{4\pi}{3}\) C: \(4 + 4\sqrt{3} + \frac{4\pi}{3}\) D: \(16 + 2\sqrt{3}  \frac{8\pi}{3}\) E: \(16  4\sqrt{3}  \frac{8\pi}{3}\) Attachment:
square.png [ 9.08 KiB  Viewed 2750 times ]
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Kindly press "+1 Kudos" to appreciate




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

Re: Square ABCD has arc AC centred at B and arc BD centred at C
[#permalink]
Show Tags
11 Sep 2014, 03:50
Took sometime, but learned some new concepts. There is always alternate ways of solution one problem.
Step 1: lets name the point of intersection of arc AC and arc BD as O. Connecting BO, we will get sector AOB with AB,OB as radius and arc AO. Similiarly connecting OC, we will get sector sector COD with CD ,OC as radius and arc DO.
Step 2. OB = OC = BC = radius of the both the circle = side of square ABCD. BOC will form a equilateral triangle.
Step 3: Area of the shaded area = Area of Square  (Area of sector AOB + Area of triangle BOC + Area of sector COD)
Step 4: Calculate area of sector AOB. angle AOB = 30° (Since BOC is equilateral triangle, all angles will be 60°). Area of sector will be (30/360) of area of circle. = (30/360)Πr² = (1/12)Π×4² = 4Π/3
Step 5: Area of equilateral triangle BOC with side r = (1/2)*(r√3/2)*r = 4√3
Step 6. Area of sector COD will be equal to area of Sector AOB (same radius, same angle).
Step 7. Area of shaded area = 16  (4Π/3 + 4√3 + 4Π/3 ) = 16  8Π/3  4√3




Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9446
Location: Pune, India

Square ABCD has arc AC centred at B and arc BD centred at C
[#permalink]
Show Tags
10 Sep 2014, 23:40
PareshGmat wrote: As shown in figure, Square ABCD has arc AC centred at B and arc BD centred at C. If AB = 4, area of shaded region is A: \(\frac{16\pi}{3}  4\sqrt{3}\) B: \(4\sqrt{3}  \frac{4\pi}{3}\) C: \(4 + 4\sqrt{3} + \frac{4\pi}{3}\) D: \(16 + 2\sqrt{3}  \frac{8\pi}{3}\) E: \(16  4\sqrt{3}  \frac{8\pi}{3}\) Experts: Please jump in. This question went top of my head. Either its too hard or too easy to make me dumb Anxiously awaiting for breakthrough.... Now that you know how to calculate the area of the other shaded region (the question you have put in the Quantitative forum), you should be able to solve this. For that solution, check: areaoftheshadedregion181249.html#p1408136Area of this shaded region = Area of square  Area of a quarter of a circle  (Area of Quarter of circle  Area of the other shaded region) Area of this shaded region = Area of square\( (1/4)*\pi*4^2  ((1/4)*\pi*4^2  (16/3)*\pi + 4*\sqrt{3})\)  Area obtained in that question if radius is 4 Area of this shaded region \(= 16  4*\pi  4*\pi + (16/3)*\pi  4*\sqrt{3}\) Area of this shaded region \(= 16  (8/3)*\pi  4\sqrt{3}\)
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1787
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)

Re: Square ABCD has arc AC centred at B and arc BD centred at C
[#permalink]
Show Tags
11 Sep 2014, 00:58
VeritasPrepKarishma wrote: PareshGmat wrote: As shown in figure, Square ABCD has arc AC centred at B and arc BD centred at C. If AB = 4, area of shaded region is Attachment: square.png A: \(\frac{16\pi}{3}  4\sqrt{3}\) B: \(4\sqrt{3}  \frac{4\pi}{3}\) C: \(4 + 4\sqrt{3} + \frac{4\pi}{3}\) D: \(16 + 2\sqrt{3}  \frac{8\pi}{3}\) E: \(16  4\sqrt{3}  \frac{8\pi}{3}\) Experts: Please jump in. This question went top of my head. Either its too hard or too easy to make me dumb Anxiously awaiting for breakthrough.... Now that you know how to calculate the area of the other shaded region (the question you have put in the Quantitative forum), you should be able to solve this. For that solution, check: areaoftheshadedregion181249.html#p1408136Area of this shaded region = Area of square  Area of a quarter of a circle  (Area of Quarter of circle  Area of the other shaded region) Area of this shaded region = Area of square\( (1/4)*\pi*4^2  ((1/4)*\pi*4^2  (16/3)*\pi + 4*\sqrt{3})\)  Area obtained in that question if radius is 4 Area of this shaded region \(= 16  4*\pi  4*\pi + (16/3)*\pi  4*\sqrt{3}\) Area of this shaded region \(= 16  (8/3)*\pi  4\sqrt{3}\) Yes Karishma, thank you so much for the calculus. That question placed was in line for this one. Can you kindly tell as to what level this question would be...?
_________________
Kindly press "+1 Kudos" to appreciate



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9446
Location: Pune, India

Re: Square ABCD has arc AC centred at B and arc BD centred at C
[#permalink]
Show Tags
11 Sep 2014, 04:21
PareshGmat wrote: VeritasPrepKarishma wrote: PareshGmat wrote: As shown in figure, Square ABCD has arc AC centred at B and arc BD centred at C. If AB = 4, area of shaded region is Attachment: square.png A: \(\frac{16\pi}{3}  4\sqrt{3}\) B: \(4\sqrt{3}  \frac{4\pi}{3}\) C: \(4 + 4\sqrt{3} + \frac{4\pi}{3}\) D: \(16 + 2\sqrt{3}  \frac{8\pi}{3}\) E: \(16  4\sqrt{3}  \frac{8\pi}{3}\) Experts: Please jump in. This question went top of my head. Either its too hard or too easy to make me dumb Anxiously awaiting for breakthrough.... Now that you know how to calculate the area of the other shaded region (the question you have put in the Quantitative forum), you should be able to solve this. For that solution, check: areaoftheshadedregion181249.html#p1408136Area of this shaded region = Area of square  Area of a quarter of a circle  (Area of Quarter of circle  Area of the other shaded region) Area of this shaded region = Area of square\( (1/4)*\pi*4^2  ((1/4)*\pi*4^2  (16/3)*\pi + 4*\sqrt{3})\)  Area obtained in that question if radius is 4 Area of this shaded region \(= 16  4*\pi  4*\pi + (16/3)*\pi  4*\sqrt{3}\) Area of this shaded region \(= 16  (8/3)*\pi  4\sqrt{3}\) Yes Karishma, thank you so much for the calculus. That question placed was in line for this one. Can you kindly tell as to what level this question would be...? This will not be a GMAT question  too convoluted and very cumbersome calculation. The first one, perhaps, is workable since it needs only one step but even that one would be 700+.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Intern
Joined: 12 May 2014
Posts: 14
Location: United States
Concentration: Strategy, Operations
GMAT Date: 10222014
GPA: 1.9
WE: Engineering (Energy and Utilities)

Re: Square ABCD has arc AC centred at B and arc BD centred at C
[#permalink]
Show Tags
11 Sep 2014, 05:04
Solved with the same method as Prasanna. I believe that most questions are solvable within time if the mind strikes in the right direction (Think that can comes with practice only)
TIP: Most geometry questions that seem unsolvable with the figure provided can be solved with some constructions. (Again comes with practice):P



Manager
Joined: 04 May 2015
Posts: 71
Concentration: Strategy, Operations
WE: Operations (Military & Defense)

Re: Square ABCD has arc AC centred at B and arc BD centred at C
[#permalink]
Show Tags
05 Sep 2015, 12:40
VeritasPrepKarishma wrote: PareshGmat wrote: As shown in figure, Square ABCD has arc AC centred at B and arc BD centred at C. If AB = 4, area of shaded region is Attachment: square.png A: \(\frac{16\pi}{3}  4\sqrt{3}\) B: \(4\sqrt{3}  \frac{4\pi}{3}\) C: \(4 + 4\sqrt{3} + \frac{4\pi}{3}\) D: \(16 + 2\sqrt{3}  \frac{8\pi}{3}\) E: \(16  4\sqrt{3}  \frac{8\pi}{3}\) Experts: Please jump in. This question went top of my head. Either its too hard or too easy to make me dumb Anxiously awaiting for breakthrough.... Now that you know how to calculate the area of the other shaded region (the question you have put in the Quantitative forum), you should be able to solve this. For that solution, check: areaoftheshadedregion181249.html#p1408136Area of this shaded region = Area of square  Area of a quarter of a circle  (Area of Quarter of circle  Area of the other shaded region) Area of this shaded region = Area of square\( (1/4)*\pi*4^2  ((1/4)*\pi*4^2  (16/3)*\pi + 4*\sqrt{3})\)  Area obtained in that question if radius is 4 Area of this shaded region \(= 16  4*\pi  4*\pi + (16/3)*\pi  4*\sqrt{3}\) Area of this shaded region \(= 16  (8/3)*\pi  4\sqrt{3}\) Hi VeritasPrepKarishmaI'm very confused I understand, calculating the square and then subtracting the area of a quarter of a circle with radius four, but I am lost after that as to how we calculate the leftover for the second arc? are you able to elaborate for me please? I had a read of the other post you linked to which asks you to calculate the opposite area. Once again, I can understand how one section is calculated, just not how this works in a simultaneous sense? Anyone else feel free to jump in too. Edit: I worked on this for a while and finally got to the bottom of it. It's about finding the difference between the arc and an equilateral triangle... My first mistake was that I did not know the formula for the area of an equilateral triangle and had to google it. Second mistake... Not being clever enough to recognize the process. Working on it though
_________________
If you found my post useful, please consider throwing me a Kudos... Every bit helps



Director
Joined: 23 Jan 2013
Posts: 542

Re: Square ABCD has arc AC centred at B and arc BD centred at C
[#permalink]
Show Tags
Updated on: 09 Sep 2015, 21:38
Option E is most favourable format in which we subtract unneeded area from 16. Answer in 1.5 min E
Originally posted by Temurkhon on 09 Sep 2015, 21:32.
Last edited by Temurkhon on 09 Sep 2015, 21:38, edited 1 time in total.



Intern
Joined: 29 Oct 2014
Posts: 35

Re: Square ABCD has arc AC centred at B and arc BD centred at C
[#permalink]
Show Tags
10 Sep 2015, 04:00
VeritasPrepKarishma wrote: PareshGmat wrote: As shown in figure, Square ABCD has arc AC centred at B and arc BD centred at C. If AB = 4, area of shaded region is Attachment: square.png A: \(\frac{16\pi}{3}  4\sqrt{3}\) B: \(4\sqrt{3}  \frac{4\pi}{3}\) C: \(4 + 4\sqrt{3} + \frac{4\pi}{3}\) D: \(16 + 2\sqrt{3}  \frac{8\pi}{3}\) E: \(16  4\sqrt{3}  \frac{8\pi}{3}\) Experts: Please jump in. This question went top of my head. Either its too hard or too easy to make me dumb Anxiously awaiting for breakthrough.... Now that you know how to calculate the area of the other shaded region (the question you have put in the Quantitative forum), you should be able to solve this. For that solution, check: areaoftheshadedregion181249.html#p1408136Area of this shaded region = Area of square  Area of a quarter of a circle  (Area of Quarter of circle  Area of the other shaded region) Area of this shaded region = Area of square\( (1/4)*\pi*4^2  ((1/4)*\pi*4^2  (16/3)*\pi + 4*\sqrt{3})\)  Area obtained in that question if radius is 4 Area of this shaded region \(= 16  4*\pi  4*\pi + (16/3)*\pi  4*\sqrt{3}\) Area of this shaded region \(= 16  (8/3)*\pi  4\sqrt{3}\) Nice explanation VeritasPrepKarishma



NonHuman User
Joined: 09 Sep 2013
Posts: 11720

Re: Square ABCD has arc AC centred at B and arc BD centred at C
[#permalink]
Show Tags
12 Aug 2018, 19:44
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: Square ABCD has arc AC centred at B and arc BD centred at C
[#permalink]
12 Aug 2018, 19:44






