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A circle is inscribed in a quadrant of a circle of radius 1 [#permalink]
 07 Mar 2010, 13:20
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A circle is inscribed in a quadrant of a circle of radius 1 unit. What is the area of the shaded region? I donot have the options and ans, but we can have discussion over this. PS: This is a difficult question and might not be tested in GMAT, but solving such questions might help us learning new concept. DISCUSSED HERE: consider-a-quarter-of-a-circle-of-radius-16-let-r-be-the-131083.html
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Last edited by gurpreetsingh on 07 Mar 2010, 14:41, edited 1 time in total.
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Re: geometry - circles [#permalink]
07 Mar 2010, 14:07
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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...!
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CEO
Status: Nothing comes easy: neither do I want.
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Re: geometry - circles [#permalink]
07 Mar 2010, 14:15
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Re: geometry - circles [#permalink]
07 Mar 2010, 14:30
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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! 
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CEO
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Re: geometry - circles [#permalink]
07 Mar 2010, 14:40
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.
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Re: geometry - circles [#permalink]
07 Mar 2010, 15:13
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
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CEO
Status: Nothing comes easy: neither do I want.
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Re: geometry - circles [#permalink]
07 Mar 2010, 21:55
Its all about team work, Gmat should also be flexible. I will find my partner to be good in verbal and then atleast i can think of 700+
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Re: geometry - circles [#permalink]
25 Apr 2010, 03:05
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Re: geometry - circles [#permalink]
25 Apr 2010, 05:32
Hussain15 wrote: One query,, how did jeteesh derive the following equation.
x+r=\sqrt{2}r
I am struck. Just look at the right angled triangle whose 2 sides are r and hypotenuse is x+r =>(x+r)^2 = r^2 +r^2 => x+r = \sqrt{2}rHere x is basically the distance from the center of bigger circle to the smaller circle.
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Re: geometry - circles [#permalink]
27 Apr 2010, 14:22
How does the first equation can turn to the second one?
2r + (\sqrt{2} - 1)r = 1
r = (\sqrt{2} - 1)
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Re: geometry - circles [#permalink]
30 Apr 2010, 12:05
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.. 
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Re: geometry - circles [#permalink]
06 May 2011, 01:46
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, 08:43, edited 2 times in total.
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Re: geometry - circles [#permalink]
06 May 2011, 07:11
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.
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Re: geometry - circles [#permalink]
01 Sep 2012, 13:14
Looking for complete solution with good explanation. I could reach half of the solution.
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Re: geometry - circles [#permalink]
01 Sep 2012, 14:11
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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Re: A circle is inscribed in a quadrant of a circle of radius 1 [#permalink]
03 Sep 2012, 05:39
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Re: A circle is inscribed in a quadrant of a circle of radius 1
[#permalink]
03 Sep 2012, 05:39
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