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Consider a quarter of a circle of radius 16. Let r be the radius of th

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Consider a quarter of a circle of radius 16. Let r be the radius of th  [#permalink]

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New post 21 Apr 2012, 09:28
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Consider a quarter of a circle of radius 16. Let r be the radius of the circle inscribed in this quarter of a circle. Find r.


A. \(16*(\sqrt{2} -1)\)

B. \(8*(\sqrt{3} -1)\)

C. \(4*(\sqrt{7} - 1)\)

D. \(12* (\sqrt{7} -1)\)

E. None of these


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Consider a quarter of a circle of radius 16. Let r be the radius of th  [#permalink]

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New post 21 Apr 2012, 11:09
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JubtaGubar wrote:
Consider a quarter of a circle of radius 16. Let r be the radius of the circle inscribed in this quarter of a circle. Find r.

A. \(16*(\sqrt{2} -1)\)

B. \(8*(\sqrt{3} -1)\)

C. \(4*(\sqrt{7} - 1)\)

D. \(12* (\sqrt{7} -1)\)

E. None of these


A question from MBA Strategy course


Look at the diagram below:
Attachment:
Quarter of circle.png
Quarter of circle.png [ 5.62 KiB | Viewed 5707 times ]

The radius of a quarter of a circle equals to the diagonal of a square made by the radii of the inscribed circle plus the radius of that circle.

Now, since the sides of a square equal to \(r\), then its diagonal equals to \(r\sqrt{2}\), hence \(r\sqrt{2}+r=16\) --> \(r=\frac{16}{\sqrt{2}+1}\).

Rationalise by multiplying both numerator and denominator by \(\sqrt{2}-1\): \(r=\frac{16(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)}\) --> apply \((a+b)(a-b)=a^2-b^2\) to the expression in the denominator: \(r=\frac{16(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)}=\frac{16(\sqrt{2}-1)}{2-1}=16(\sqrt{2}-1)\).

Answer: A.

Hope it's clear.
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Re: Consider a quarter of a circle of radius 16. Let r be the radius of th  [#permalink]

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New post 23 Apr 2012, 07:13
Thank you Bunuel!

I got 16/(sqr2 +1) and just forgot to multiply by (sqr2 -1).
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Re: A circle is inscribed in a quadrant of a circle of radius 1  [#permalink]

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Re: A circle is inscribed in a quadrant of a circle of radius 1   [#permalink] 30 Jan 2019, 01:10
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