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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 28427 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, 07:43, 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.
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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. \((\pi/4-\pi*(\sqrt{2} - 1)^2-((\sqrt{2} - 1)^2-\pi*(\sqrt{2} - 1)^2/4))/2 = 0.10478\)