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 $a+b+c$.
Rangpur Secondary (Higher Secondary) 2011/10
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- Tahmid Hasan
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Re: Rangpur Secondary 2011/10
Diagram uploaded.
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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