1970 AHSME Problems/Problem 24
Problem
An equilateral triangle and a regular hexagon have equal perimeters. If the area of the triangle is , then the area of the hexagon is
Solution 1
Let be our regular hexagon, with centre - and join and . Note that we form six equilateral triangles with sidelength , where is the sidelength of the triangle (since the perimeter of the two polygons are equal). If the area of the original equilateral triangle is , then the area of each of the six smaller triangles is (by similarity area ratios).
Thus, the area of the hexagon is , hence our answer is
See also
1970 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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