Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 19"
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− | <math>A \Delta B</math> is a [[right triangle]] with an angle of <math>45^{\circ}</math>, so it is a <math>45-45-90</math> triangle with <math> | + | <math>A \Delta B</math> is a [[right triangle]] with an angle of <math>45^{\circ}</math>, so it is a <math>45-45-90</math> triangle with <math>B\Delta = \frac{AB}{\sqrt{2}} = 1</math>. |
The area of the entire circle is <math>(1)^2\pi = \pi</math>. The central angle of the sector is <math>\frac{180-45}{2} = \frac{135}{2}</math>, so the area is <math>\frac{\frac{135}{2}}{360} = \frac{3}{16}\pi</math>. | The area of the entire circle is <math>(1)^2\pi = \pi</math>. The central angle of the sector is <math>\frac{180-45}{2} = \frac{135}{2}</math>, so the area is <math>\frac{\frac{135}{2}}{360} = \frac{3}{16}\pi</math>. |
Revision as of 08:14, 12 August 2008
Problem
In the figure, is an isosceles triangle with and $\ang A=45^\circ$ (Error compiling LaTeX. Unknown error_msg). If is an altitude of the triangle and the sector belongs to the circle , the area of the shaded region is
Solution
is a right triangle with an angle of , so it is a triangle with .
The area of the entire circle is . The central angle of the sector is , so the area is .
The area of the entire triangle is . Thus, the answer is .
See also
2006 Cyprus MO, Lyceum (Problems) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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