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 Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

I donot have the options and ans, but we can have discussion over this.

Wild Spin!!!

I am able to take out the area of the inscribed circle... after that am lost...! _________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

one of my teacher once told me, 2 fools can solve any problem...lol

hahaha... :D

Let the radius of the smaller circle be r... and let the distance from center of bigger circle to the circumference of inscribed circle be x. Check the figure.

2r + x = 1

x + r = \(\sqrt{2}\)r Therefore x = (\(\sqrt{2} - 1\))r

Therefore 2r + (\(\sqrt{2} - 1\))r = 1 Which gives r = (\(\sqrt{2} - 1\))

Now you can find the area of the smaller circle....

This is the limit, this fool can reach! All over to the other fool!

Attachments

Circle.png [ 16.19 KiB | Viewed 16572 times ]

_________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

Shaded portion = Area of sector 1 - area of triangle 1 - area of sector 2

where area of sector 1 = area made by 45 degree of bigger circle.

area of triangle 1 is where you have applied Pythagoras theorem

area of sector 2 = area made by 135 degree of smaller circle.

Cool!!!! Do they allow two fools together in GMAT :lol _________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

if v can extend the figure like take big circle ... then put in 4 circles of (root2 -1) radius in it.... then joining the centres of four circles v will form a square with an area = (2(root2 -1))^2 coz radiu smaller circles is root2 -1 ..... then to get the shaded region

= {area of big circle - {3(area of smaller circle) + (2(root2 -1))^2}} / 8 ...... v subtracted three circles coz i circle ia in the square..

area of small circle is=\(\pi{r^2}\)=\(\pi*(\sqrt{2} - 1)^2\) area of Quarter=\((\pi*{1^2})/4=\pi/4\). area of shaded region=\((\pi/4-\pi*(\sqrt{2} - 1)^2 -(r^2-1/4(\pi*(\sqrt{2} - 1)^2)) / 2\)

Last edited by annmary on 06 May 2011, 08:43, edited 2 times in total.

the bigger radius comes out to be = 1 + 2^(1/2). So from the sector Area,subtract the circle area and the area of the sector between circle and edge of the bigger sector

the area of the edge sector = 1- pi/4

thus the shaded region area can be found out. _________________

area of Quarter of big circle=\((\pi*{1^2})/4=\pi/4\). The radius r of the inscribed smaller circle: \(r = \sqrt{2} - 1\) See explanation for this above.

Area not covered by small circle = \(\pi/4-\pi*(\sqrt{2} - 1)^2\)

Small area in the corner (near center of big circle) not covered by small circle: Small square (see picture above) - a quarter of the smaller circle \((\sqrt{2} - 1)^2-\pi*(\sqrt{2} - 1)^2/4\)

Area not covered by small circle and not including the small area in the corner: \(\pi/4-\pi*(\sqrt{2} - 1)^2-((\sqrt{2} - 1)^2-\pi*(\sqrt{2} - 1)^2/4)\)

This area now represents the shaded area upper left and its mirror area bottom right. So we only have to divide by 2 to get the area of the shaded area. \((\pi/4-\pi*(\sqrt{2} - 1)^2-((\sqrt{2} - 1)^2-\pi*(\sqrt{2} - 1)^2/4))/2 = 0.10478\)

So, my final tally is in. I applied to three b schools in total this season: INSEAD – admitted MIT Sloan – admitted Wharton – waitlisted and dinged No...

A few weeks ago, the following tweet popped up in my timeline. thanks @Uber_Mumbai for showing me what #daylightrobbery means!I know I have a choice not to use it...

“This elective will be most relevant to learn innovative methodologies in digital marketing in a place which is the origin for major marketing companies.” This was the crux...