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

 It is currently 15 Nov 2018, 17:44

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

Events & Promotions

Events & Promotions in November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
• Free GMAT Strategy Webinar

November 17, 2018

November 17, 2018

07:00 AM PST

09:00 AM PST

Nov. 17, 7 AM PST. Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.
• GMATbuster's Weekly GMAT Quant Quiz # 9

November 17, 2018

November 17, 2018

09:00 AM PST

11:00 AM PST

Join the Quiz Saturday November 17th, 9 AM PST. The Quiz will last approximately 2 hours. Make sure you are on time or you will be at a disadvantage.

In the figure, AB is parallel to CD and AB=CD=2 2 . If the

Author Message
Manager
Joined: 15 Feb 2011
Posts: 205
In the figure, AB is parallel to CD and AB=CD=2 2 . If the  [#permalink]

Show Tags

15 Aug 2011, 05:30
00:00

Difficulty:

(N/A)

Question Stats:

67% (02:10) correct 33% (01:34) wrong based on 9 sessions

HideShow timer Statistics

In the figure, AB is parallel to CD and $$AB=CD=2 \sqrt{2}$$. If the distance between $$AB$$ and $$CD$$ is $$2\sqrt{2}$$, what is the area of the shaded region?

(A) 2+pi
(B) 2+2pi
(C) 4+pi
(D) 4+2pi
(E) 4+4pi

Attachment:

geometry_why_not.JPG [ 6.7 KiB | Viewed 4585 times ]

--== Message from the GMAT Club Team ==--

THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION.
This discussion does not meet community quality standards. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.

Attachments

Geo not again.docx [13.7 KiB]

Manager
Status: Quant 50+?
Joined: 02 Feb 2011
Posts: 97
Concentration: Strategy, Finance
Schools: Tuck '16, Darden '16

Show Tags

15 Aug 2011, 05:46
Find length AD with pythagorean theorem.

Area full circle = 4pi

Area full square = (2root2)*(2root2)= 8

Now area of the shaded region is the area of the full circle, minus 1/2 the difference between the area of the circle and the area of the inscribed square.

4pi - 1/2(4pi-8) = 4pi-2pi+4 = 2pi + 4 = D
Manager
Joined: 15 Feb 2011
Posts: 205

Show Tags

15 Aug 2011, 06:04
eragotte wrote:
Find length AD with pythagorean theorem.

Area full circle = 4pi

Area full square = (2root2)*(2root2)= 8

Now area of the shaded region is the area of the full circle, minus 1/2 the difference between the area of the circle and the area of the inscribed square.

4pi - 1/2(4pi-8) = 4pi-2pi+4 = 2pi + 4 = D

hmmmm but why did you minus 1/2??
Manager
Status: Quant 50+?
Joined: 02 Feb 2011
Posts: 97
Concentration: Strategy, Finance
Schools: Tuck '16, Darden '16

Show Tags

15 Aug 2011, 06:10
DeeptiM wrote:
eragotte wrote:
Find length AD with pythagorean theorem.

Area full circle = 4pi

Area full square = (2root2)*(2root2)= 8

Now area of the shaded region is the area of the full circle, minus 1/2 the difference between the area of the circle and the area of the inscribed square.

4pi - 1/2(4pi-8) = 4pi-2pi+4 = 2pi + 4 = D

hmmmm but why did you minus 1/2??

I am minusing 1/2 of the difference between the area of the circle and the area of the square. Think about it and you should see the the midpoint between the area of the circle and the area of the square is the area of the shaded region. Not really sure how to better explain it. Another way to do it is area of square + area of circle divided by two. Actually that is probably easier.

(4pi + 8)/2 = 2pi + 4
Retired Moderator
Joined: 20 Dec 2010
Posts: 1829

Show Tags

15 Aug 2011, 06:37
1
DeeptiM wrote:
In the figure, AB is parallel to CD and $$AB=CD=2 \sqrt{2}$$. If the distance between $$AB$$ and $$CD$$ is $$2\sqrt{2}$$, what is the area of the shaded region?

(A) 2+pi
(B) 2+2pi
(C) 4+pi
(D) 4+2pi
(E) 4+4pi

Attachment:
The attachment geometry_why_not.JPG is no longer available

$$ABCD$$ is a square with $$Side=2\sqrt{2}$$

Area of square$$=(2\sqrt{2})^2=4*2=8$$

Let's say we join the diagonals of the square $$AD$$ and $$BC$$, intersecting at point O.

The four triangular regions(shown with greenish shades) within the square, viz $$AOB, BOD, DOC, COA$$ will have equal areas each of 2 units i.e. 8/4=2

$$\triangle ACD$$ is a right-angled triangle.

Now, in order to find the area of the segment AC(orange area on left), we just need to find the area of the sector $$AOC$$ (Dark Orange+ Darkest Green) and subtract the area of triangle $$AOC=2$$ (Dark Green)

$$Area \hspace{2} of \hspace{2} sector=\frac{\theta}{360}*\pi*r^2$$

$$Area \hspace{2} of \hspace{2} sector=\frac{90}{360}*\pi*2^2=\pi$$ {:Note: theta=AOC=90. Square's diagonals make $$90^{\circ}$$ angle at intersection.}

Area of segment= Area of sector-Area of quarter of the square$$= \pi - 2$$

The shaded portion has two segments(Left Orange+Right Orange) and the square(sum of all green regions).

Total Area=2*Area of segment+Area of square$$=2(\pi -2)+8=2\pi-4+8=2\pi+4$$

Ans: "D"
Attachments

Square_Inscribed_In_Circle.JPG [ 17.9 KiB | Viewed 4544 times ]

_________________
Manager
Joined: 07 Jun 2011
Posts: 51

Show Tags

15 Aug 2011, 09:59
From flukes diagram the green and the bleueish green forms one quarter of the area of the whole circle. This is because all the angles are equal.

so calculate the area of the complete circle and divide that value by 4 .

We are now left with 2 isosceles triangles for which we are to find the area. Very straight forward because we have the value of the base and hight.
Senior Manager
Joined: 03 Mar 2010
Posts: 379
Schools: Simon '16 (M)

Show Tags

15 Aug 2011, 12:01
ABCD is a square with side $$2\sqrt{2}$$
BC and AD will be diagonals which are perpendicular bisector at center 0 of circle.
Imagine 0 as center of circle.
Area of sector A0C = Area of sector B0D= Central angle/360 * pi r^2 = 90/360*pi*4=pi
Area of triangle 0CD = Area of triangle 0AB= 1/2 * 2 * 2 =2
Area of shaded region = Area of sector A0C + Area of sector B0D+Area of triangle 0CD + Area of triangle 0AB
= pi+pi+2+2=4+2pi

OA D.
_________________

My dad once said to me: Son, nothing succeeds like success.

Manager
Joined: 15 Feb 2011
Posts: 205

Show Tags

15 Aug 2011, 18:38
Awsum guys, thnx a ton for the explanation..
Manager
Joined: 27 May 2008
Posts: 105

Show Tags

17 Aug 2011, 02:52
Nice explanation fluke..
VP
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 1041

Show Tags

17 Aug 2011, 03:06
joining A-D and B-C we have 4 triangles.

area of top and bottom triangles = 2 * 1/2 * (base=2*2^(1/2)) * (altitude = 2^(1/2)

altitude = 2^(1/2) because distance is split equally at the center = [2* 2^(1/2)]/2

hence area = 4.

now area of left and right triangles = 2* [angle/360] * pi * r^2

r^2 = 4 hence r=2 also,
angles for the left triangle = 45-45-90 (center)

thus, 90/360 * pi * 4 * 2 = 2pi

thus 4+ 2pi. D
Manager
Joined: 28 May 2011
Posts: 158
Location: United States
GMAT 1: 720 Q49 V38
GPA: 3.6
WE: Project Management (Computer Software)

Show Tags

17 Aug 2011, 09:28
People have already done lot of hard work in explaining the solution, So agreeing to them I would just give my answer & that is D

--== Message from the GMAT Club Team ==--

THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION.
This discussion does not meet community quality standards. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.

_________________

-------------------------------------------------------------------------------------------------------------------------------
http://gmatclub.com/forum/a-guide-to-the-official-guide-13-for-gmat-review-134210.html
-------------------------------------------------------------------------------------------------------------------------------

Non-Human User
Joined: 09 Sep 2013
Posts: 8776
Re: In the figure, AB is parallel to CD and AB=CD=2 2 . If the  [#permalink]

Show Tags

02 Nov 2018, 10:10
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: In the figure, AB is parallel to CD and AB=CD=2 2 . If the &nbs [#permalink] 02 Nov 2018, 10:10
Display posts from previous: Sort by