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# A circle is inscribed in a quadrant of a circle of radius 1

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CEO
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07 Mar 2010, 13:20
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67% (01:11) correct 33% (01:13) wrong based on 2 sessions

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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.

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Last edited by gurpreetsingh on 07 Mar 2010, 14:41, edited 1 time in total.
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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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Kudos [?]: 1514 [1] , given: 235

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07 Mar 2010, 14:15
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Shar your steps, toghther we might solve.

one of my teacher once told me, 2 fools can solve any problem...lol
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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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Circle.png [ 16.19 KiB | Viewed 18041 times ]

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CEO
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Kudos [?]: 1514 [0], given: 235

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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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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
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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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25 Apr 2010, 03:05
One query,, how did jeteesh derive the following equation.

$$x+r=\sqrt{2}r$$

I am struck.
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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

=> $$x+r = \sqrt{2}r$$

Here x is basically the distance from the center of bigger circle to the smaller circle.
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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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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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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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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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01 Sep 2012, 13:14
Looking for complete solution with good explanation. I could reach half of the solution.
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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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03 Sep 2012, 05:39
Re: A circle is inscribed in a quadrant of a circle of radius 1   [#permalink] 03 Sep 2012, 05:39
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