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As shown in figure, square ABCD has arc BPD centred at C and arc BQD [#permalink]
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10 Sep 2014, 02:21
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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 3573 times ]
A: \(16  4\pi\) B: \(8  2\pi\) C: \(8\pi  16\) D: \(4\pi  8\) E: \(2\pi  4\)
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Re: As shown in figure, square ABCD has arc BPD centred at C and arc BQD [#permalink]
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10 Sep 2014, 04:40
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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 3153 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.
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Re: As shown in figure, square ABCD has arc BPD centred at C and arc BQD [#permalink]
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10 Sep 2014, 06:40
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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.



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Re: As shown in figure, square ABCD has arc BPD centred at C and arc BQD [#permalink]
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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.



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Re: As shown in figure, square ABCD has arc BPD centred at C and arc BQD [#permalink]
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10 Sep 2014, 07:47
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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 3106 times ]
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Re: As shown in figure, square ABCD has arc BPD centred at C and arc BQD [#permalink]
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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.



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Re: As shown in figure, square ABCD has arc BPD centred at C and arc BQD [#permalink]
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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 2862 times ]
Consider any Quarter circle \(Its Area = \frac{\pi4^2}{4} = 4x + 8  2x\) \(x = 2\pi  4\) Answer = E
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Re: As shown in figure, square ABCD has arc BPD centred at C and arc BQD [#permalink]
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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



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As shown in figure, square ABCD has arc BPD centred at C and arc BQD [#permalink]
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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



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