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PareshGmat
Joined: 27 Dec 2012
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
60% (02:44) correct
40%
(02:57)
wrong
based on 120
sessions
History
Date
Time
Result
Not Attempted Yet
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: \(16 - 4\sqrt{3} - \frac{8\pi}{3}\)
square.png [ 9.08 KiB | Viewed 11089 times ]
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 11089 times ]
nvijayprasanna
Joined: 17 Apr 2012
Last visit: 10 May 2017
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Location: United States
WE:Information Technology (Computer Software)
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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
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
General Discussion
PareshGmat
Make the complete diagram to understand the relevance of the point where the two arcs intersect.
Attachment:
Ques4.jpg [ 20.02 KiB | Viewed 9974 times ]
The area you are looking for is enclosed between arc AC, arc BC and line BA. ABC is an equilateral triangle. So angle ABC is 60
Shaded region \(= (1/6)*\pi*r^2 + ((1/6)*\pi*r^2 - \sqrt{3}r^2/4)\)
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PareshGmat
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: area-of-the-shaded-region-181249.html#p1408136
Area 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 \(= 16 - (8/3)*\pi - 4\sqrt{3}\)
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PareshGmat
Joined: 27 Dec 2012
Last visit: 10 Jul 2016
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VeritasPrepKarishma
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...?
PareshGmat
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+.
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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
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
hi ,
will this type of question appear on GMAT?
also i was unable to understand the solution , can you please explain this in detail.
will this type of question appear on GMAT?
also i was unable to understand the solution , can you please explain this in detail.
Jerry1982
Try making the diagram from the given diagram. You are given that "Square ABCD has arc AC centred at B & arc BD centred at C". So make the circle centered at B and the circle centered at C. Now, remove the lines of the square that you don't care about and focus on only the shaded area. Don't worry about the way I have named the vertices - they are not according to the given diagram. Just note that the shaded portion corresponds to the area bounded by arc AC, arc BC and line BA in my diagram. Now join the points that define the shaded region to get an equilateral triangle (they are all radii of circles with equal radii).
Then you get that the angle subtended at the center is 60 degrees which means area of the sector is 1/6th (which is 60/360) of the area of the circle. But that is not all the shaded region. There is some extra shaded region lying between arc BC and line BC (in my diagram). To get the value of this, we subtract the area of the triangle BAC from the area of the sector BAC (in my diagram)
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DropBear
Joined: 04 May 2015
Last visit: 30 Jul 2016
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Concentration: Strategy, Operations
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VeritasPrepKarishma
Hi VeritasPrepKarishma
I'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.
Moderator Note: 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 
Schools: Cambridge'16
Posts: 405
Option E is most favourable format in which we subtract unneeded area from 16. Answer in 1.5 min 
E
E
PareshGmat
Area Shaded = Area Square ABCD - Area Equilateral BOC - 2(Area of Sector AOB or DOC)
Area Square = 16
Area Equilateral = \(x^2\sqrt{3}/4=16\sqrt{3}/4=4\sqrt{3}\)
Area Sector = Area Circle * 30/360 = \(4^2π*30/360=16π/12=4π/3\)
Area Shaded = \(16 - 4\sqrt{3} - 2(4π/3) = 16 - 4\sqrt{3} - 8π/3\)
Answer (E)
Attachments
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