Rangpur Secondary (Higher Secondary) 2011/10

Problem for Secondary Group from Divisional Mathematical Olympiad will be solved here.
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Moon
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Rangpur Secondary (Higher Secondary) 2011/10

Unread post by Moon » Wed Feb 02, 2011 6:50 pm

Problem 10:
$ABCD$ is a rectangle where $AB = 3$, $BC = 6$ and $CD=CE$. The area of the part of the quadrangle $CDGF$ that lies outside the circular arc can be expressed as $a-b \sqrt{3}-\frac{c\cdot \pi}{4}$ where $a$, $b$, $c$ are integers. Find $a+b+c$.
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Tahmid Hasan
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Re: Rangpur Secondary 2011/10

Unread post by Tahmid Hasan » Wed Feb 02, 2011 8:07 pm

where is the point $E$ situated?
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Re: Rangpur Secondary 2011/10

Unread post by Moon » Wed Feb 02, 2011 8:53 pm

Diagram uploaded. :)
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Mehfuj Zahir
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Re: Rangpur Secondary (Higher Secondary) 2011/10

Unread post by Mehfuj Zahir » Thu Feb 03, 2011 12:16 pm

Connect the point C,G.Find the area of triangle CGF=9.SO,BG=6.Then using pythagoreous find BF&get GF.Now the area of triangle CGF&CDG is equal.Find area of quadrangle DGFC.The angleCDF is 30.Then subtract area inscribed by the arch DF from quadrangle DGFC.Ans should be a+b+C=30

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