## 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

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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"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

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Tahmid Hasan
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### Re: Rangpur Secondary 2011/10

where is the point $E$ situated?
বড় ভালবাসি তোমায়,মা

Moon
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### Re: Rangpur Secondary 2011/10

Diagram uploaded. "Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

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Mehfuj Zahir
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### Re: Rangpur Secondary (Higher Secondary) 2011/10

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