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Re: geometry - circles [#permalink]
07 Mar 2010, 13:07
1
This post received KUDOS
gurpreetsingh wrote:
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|
Re: geometry - circles [#permalink]
07 Mar 2010, 13:30
2
This post received KUDOS
gurpreetsingh wrote:
Shar your steps, toghther we might solve.
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 15463 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|
Re: geometry - circles [#permalink]
07 Mar 2010, 14:13
gurpreetsingh wrote:
lol i think its solved.
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..
Re: geometry - circles [#permalink]
06 May 2011, 00:46
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.
Re: geometry - circles [#permalink]
06 May 2011, 06:11
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. \((\pi/4-\pi*(\sqrt{2} - 1)^2-((\sqrt{2} - 1)^2-\pi*(\sqrt{2} - 1)^2/4))/2 = 0.10478\)
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