1970 AHSME Problems/Problem 24

Problem

An equilateral triangle and a regular hexagon have equal perimeters. If the area of the triangle is $2$ , then the area of the hexagon is

$\text{(A) } 2\quad \text{(B) } 3\quad \text{(C) } 4\quad \text{(D) } 6\quad \text{(E) } 12$

Solution 1

Let $ABCDEF$ be our regular hexagon, with centre $O$ - and join $AO, BO, CO, DO, EO,$ and $FO$ . Note that we form six equilateral triangles with sidelength $\frac{s}{2}$ , where $s$ is the sidelength of the triangle (since the perimeter of the two polygons are equal.) If the area of the original equilateral triangle is $2$ , then the area of each of the six smaller triangles is $1/4\times 2 = \frac{1}{2}$ (by similarity area ratios).

Thus, the area of the hexagon is $6\times \frac{1}{2}=3$ , hence our answer is

1970 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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1970 AHSME Problems/Problem 24

Problem

Solution 1

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