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

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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: $$2\pi - 4$$
[Reveal] Spoiler: OA

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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 = $$4^2-2*(16 - 4\pi)=8\pi-16$$.

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

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 = $$4^2-2*(16 - 4\pi)=8\pi-16$$.

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

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 = $$4^2-2*(16 - 4\pi)=8\pi-16$$.

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

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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Joined: 27 Dec 2012
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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

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

$$x = 2\pi - 4$$

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

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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07 Apr 2018, 08:28
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Re: As shown in figure, square ABCD has arc BPD centred at C and arc BQD   [#permalink] 07 Apr 2018, 08:28
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