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# Square ABCD has arc AC centred at B and arc BD centred at C

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Square ABCD has arc AC centred at B and arc BD centred at C  [#permalink]

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10 Sep 2014, 22:22
2
10
00:00

Difficulty:

65% (hard)

Question Stats:

62% (02:41) correct 38% (02:51) wrong based on 112 sessions

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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 ]

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Re: Square ABCD has arc AC centred at B and arc BD centred at C  [#permalink]

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11 Sep 2014, 03:50
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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
##### General Discussion
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Square ABCD has arc AC centred at B and arc BD centred at C  [#permalink]

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10 Sep 2014, 23:40
2
1
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.

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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Re: Square ABCD has arc AC centred at B and arc BD centred at C  [#permalink]

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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.

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 $$= 16 - (8/3)*\pi - 4\sqrt{3}$$

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...?
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Re: Square ABCD has arc AC centred at B and arc BD centred at C  [#permalink]

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11 Sep 2014, 04:21
1
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.

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 $$= 16 - (8/3)*\pi - 4\sqrt{3}$$

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+.
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Re: Square ABCD has arc AC centred at B and arc BD centred at C  [#permalink]

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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
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Re: Square ABCD has arc AC centred at B and arc BD centred at C  [#permalink]

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05 Sep 2015, 12:40
1
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.

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 $$= 16 - (8/3)*\pi - 4\sqrt{3}$$

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.

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

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Re: Square ABCD has arc AC centred at B and arc BD centred at C  [#permalink]

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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.
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Re: Square ABCD has arc AC centred at B and arc BD centred at C  [#permalink]

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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.

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 $$= 16 - (8/3)*\pi - 4\sqrt{3}$$

Nice explanationVeritasPrepKarishma
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Re: Square ABCD has arc AC centred at B and arc BD centred at C  [#permalink]

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