1979 AHSME Problems/Problem 10

Problem 10

If $P_1P_2P_3P_4P_5P_6$ is a regular hexagon whose apothem (distance from the center to midpoint of a side) is $2$ , and $Q_i$ is the midpoint of side $P_iP_{i+1}$ for $i=1,2,3,4$ , then the area of quadrilateral $Q_1Q_2Q_3Q_4$ is

$\textbf{(A) }6\qquad \textbf{(B) }2\sqrt{6}\qquad \textbf{(C) }\frac{8\sqrt{3}}{3}\qquad \textbf{(D) }3\sqrt{3}\qquad \textbf{(E) }4\sqrt{3}$

Solution

Solution by e_power_pi_times_i

Notice that quadrilateral $Q_1Q_2Q_3Q_4$ consists of $3$ equilateral triangles with side length $2$ . Thus the area of the quadrilateral is $3\cdot(\frac{2^2\cdot\sqrt{3}}{4}) = 3\cdot\sqrt{3} = \boxed{\textbf{(D) } 3\sqrt{3}}$ .

1979 AHSME (Problems • Answer Key • Resources)
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1979 AHSME Problems/Problem 10

Problem 10

Solution

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