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 $ab \sqrt{3}\frac{c\cdot \pi}{4}$ where $a$, $b$, $c$ are integers. Find $a+b+c$.
Rangpur Secondary (Higher Secondary) 2011/10
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Please don't post problems (by starting a topic) in the "Secondary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
Rangpur Secondary (Higher Secondary) 2011/10
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 Tahmid Hasan
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Re: Rangpur Secondary 2011/10
Diagram uploaded.
"Inspiration is needed in geometry, just as much as in poetry."  Aleksandr Pushkin
Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.
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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