1999 AHSME Problems/Problem 23
Problem
The equiangular convex hexagon has and The area of the hexagon is
Solution
Solution 1
Equiangularity means that each internal angle must be exactly . The information given by the problem statement looks as follows:
We can now place this incomplete polygon onto a triangular grid, finish it, compute its area in unit triangles, and multiply the result by the area of the unit triangle.
We see that the figure contains unit triangles, and therefore its area is .
Solution 2
An equiangular hexagon can be made by drawing an equilateral triangle and cutting out smaller triangles from the corners. Labeling the triangle and and drawing of length one will remove one equilateral triangle of side length , and drawing will take out another equilateral triangle of side length .Labeling the other sides of the smaller equilateral triangles, we can find that , or the side length of the equilateral triangle is . Now, because we know what the side length of the triangle is, what is, and it is given that is , we can find the length of , . Now, to calculate the area of the hexagon we can simply subtract the area of the smaller equilateral triangles from the larger equilateral triangle. The areas of the smaller equilateral triangles are , and and the area of the large equilateral triangle is so the area of the hexagon would be
Solution 2
An equiangular hexagon can be made by drawing an equilateral triangle and cutting out smaller triangles from the corners. Labeling the triangle and and drawing of length one will remove one equilateral triangle of side length , and drawing will take out another equilateral triangle of side length .Labeling the other sides of the smaller equilateral triangles, we can find that , or the side length of the equilateral triangle is . Now, because we know what the side length of the triangle is, what is, and it is given that is , we can find the length of , . Now, to calculate the area of the hexagon we can simply subtract the area of the smaller equilateral triangles from the larger equilateral triangle. The areas of the smaller equilateral triangles are , and and the area of the large equilateral triangle is so the area of the hexagon would be
See also
1999 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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