Geometry 6 : GMAT Problem Solving (PS)
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# Geometry 6

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02 Oct 2009, 08:47
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Q.
If ABC is a quater circle and a circle is inscribed in it and if AB= 1 cm, Find radius of inner circle.

a.2^1/2- 1
b. (\sqrt{2}+1)/2
c. \sqrt{2}- 1/2
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GEOMETRY 6.JPG [ 9.99 KiB | Viewed 1791 times ]

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04 Oct 2009, 15:24

Diagonal of a square made of r=r*\sqrt{2}

R=1= r*\sqrt{2}+r --> r=1/(\sqrt{2}+1)=2^1/2-1

A.
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01 Nov 2010, 02:03
Bunuel, could you please explain this:

Quote:
R=1= r*\sqrt{2}+r --> r=1/(\sqrt{2}+1)=2^1/2-1
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24 Dec 2010, 07:59
Bunuel wrote:

Diagonal of a square made of r=r*\sqrt{2}

R=1= r*\sqrt{2}+r --> r=1/(\sqrt{2}+1)=2^1/2-1

A.

Colud not understand this.. can someone help ?

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24 Dec 2010, 08:22
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nonameee wrote:
Bunuel, could you please explain this:

Quote:
R=1= r*\sqrt{2}+r --> r=1/(\sqrt{2}+1)=2^1/2-1

hirendhanak wrote:
Bunuel wrote:

Diagonal of a square made of r=r*\sqrt{2}

R=1= r*\sqrt{2}+r --> r=1/(\sqrt{2}+1)=2^1/2-1

A.

Colud not understand this.. can someone help ?

Look at the diagram:
Attachment:

GEOMETRY 6.PNG [ 11.76 KiB | Viewed 1310 times ]
Diagonal BO of the square made by the radii of the small circle equals to $$BO=\sqrt{r^2+r^2}=r*\sqrt{2}$$;

Now, the radius of the big circle BD equals to $$BO+OD=r*\sqrt{2}+r=1$$ --> $$r=\frac{1}{\sqrt{2}+1}=\sqrt{2}-1$$

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25 Dec 2010, 09:35
Nice explanation.
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25 Dec 2010, 15:53
Oh this one was tough.
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Re: Geometry 6   [#permalink] 25 Dec 2010, 15:53
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