Difference between revisions of "2006 AIME II Problems/Problem 1"
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2116(\sqrt2+1)&=[ABCDEF]\\ | 2116(\sqrt2+1)&=[ABCDEF]\\ | ||
&=2\cdot {1\over2}x^2+x\cdot x\sqrt2=x^2(1+\sqrt2), | &=2\cdot {1\over2}x^2+x\cdot x\sqrt2=x^2(1+\sqrt2), | ||
− | \end{align*}</cmath>so <math>x^2=2116</math>, and <math>x=\boxed{ | + | \end{align*}</cmath>so <math>x^2=2116</math>, and <math>x=\boxed{046}</math>. |
<asy> | <asy> |
Latest revision as of 20:38, 22 December 2021
Contents
Problem
In convex hexagon , all six sides are congruent, and are right angles, and and are congruent. The area of the hexagonal region is Find .
Solution
Let the side length be called , so .
The diagonal . Then the areas of the triangles AFB and CDE in total are , and the area of the rectangle BCEF equals
Then we have to solve the equation
.
Therefore, is .
Solution 2
Because , , , and are congruent, the degree-measure of each of them is . Lines and divide the hexagonal region into two right triangles and a rectangle. Let . Then . Thus so , and .
See also
2006 AIME II (Problems • Answer Key • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.