GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 24 May 2019, 22:52

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1812
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
As shown in figure, square ABCD has arc BPD centred at C and arc BQD  [#permalink]

### Show Tags

10 Sep 2014, 02:21
3
8
00:00

Difficulty:

55% (hard)

Question Stats:

69% (02:26) correct 31% (02:59) wrong based on 165 sessions

### HideShow timer Statistics

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 5478 times ]

A: $$16 - 4\pi$$

B: $$8 - 2\pi$$

C: $$8\pi - 16$$

D: $$4\pi - 8$$

E: $$2\pi - 4$$

_________________
Kindly press "+1 Kudos" to appreciate
Math Expert
Joined: 02 Sep 2009
Posts: 55271
Re: As shown in figure, square ABCD has arc BPD centred at C and arc BQD  [#permalink]

### Show Tags

10 Sep 2014, 04:40
2
3
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 4642 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.
_________________
##### General Discussion
Intern
Joined: 17 Apr 2012
Posts: 14
Location: United States
WE: Information Technology (Computer Software)
Re: As shown in figure, square ABCD has arc BPD centred at C and arc BQD  [#permalink]

### Show Tags

10 Sep 2014, 06:40
1
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.
Manager
Joined: 21 Sep 2012
Posts: 213
Location: United States
Concentration: Finance, Economics
Schools: CBS '17
GPA: 4
WE: General Management (Consumer Products)
Re: As shown in figure, square ABCD has arc BPD centred at C and arc BQD  [#permalink]

### Show Tags

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.
Math Expert
Joined: 02 Sep 2009
Posts: 55271
Re: As shown in figure, square ABCD has arc BPD centred at C and arc BQD  [#permalink]

### Show Tags

10 Sep 2014, 07:47
1
1
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 4594 times ]

_________________
Manager
Joined: 21 Sep 2012
Posts: 213
Location: United States
Concentration: Finance, Economics
Schools: CBS '17
GPA: 4
WE: General Management (Consumer Products)
Re: As shown in figure, square ABCD has arc BPD centred at C and arc BQD  [#permalink]

### Show Tags

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.
SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1812
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: As shown in figure, square ABCD has arc BPD centred at C and arc BQD  [#permalink]

### Show Tags

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 4349 times ]

Consider any Quarter circle

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

$$x = 2\pi - 4$$

_________________
Kindly press "+1 Kudos" to appreciate
Intern
Joined: 28 Dec 2015
Posts: 39
Re: As shown in figure, square ABCD has arc BPD centred at C and arc BQD  [#permalink]

### Show Tags

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

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
Intern
Joined: 15 Jul 2013
Posts: 2
As shown in figure, square ABCD has arc BPD centred at C and arc BQD  [#permalink]

### Show Tags

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
Current Student
Joined: 18 Oct 2014
Posts: 834
Location: United States
GMAT 1: 660 Q49 V31
GPA: 3.98
Re: As shown in figure, square ABCD has arc BPD centred at C and arc BQD  [#permalink]

### Show Tags

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

_________________
I welcome critical analysis of my post!! That will help me reach 700+
Non-Human User
Joined: 09 Sep 2013
Posts: 11012
Re: As shown in figure, square ABCD has arc BPD centred at C and arc BQD  [#permalink]

### Show Tags

07 Apr 2018, 08:28
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: As shown in figure, square ABCD has arc BPD centred at C and arc BQD   [#permalink] 07 Apr 2018, 08:28
Display posts from previous: Sort by