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1999 AHSME Problems/Problem 23

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Problem

The equiangular convex hexagon ABCDEF has AB = 1, BC = 4, CD = 2, and DE = 4. The area of the hexagon is \mathrm{(A) \ } \frac {15}2\sqrt{3} \qquad \mathrm{(B) \ }9\sqrt{3} \qquad \mathrm{(C) \ }16 \qquad \mathrm{(D) \ }\frac{39}4...

Solution

Equiangularity means that each internal angle must be exactly 120^\circ. The information given by the problem statement looks as follows:

unitsize(0.5cm);pair O=(0,0), E=dir(0), NE=dir(60), NW=dir(120);draw(O -- (O+E) -- (O+E+4*NE) -- (O+E+4*NE+2*NW) -- (O-3*E+4*...

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.

unitsize(0.5cm);pair O=(0,0), E=dir(0), NE=dir(60), NW=dir(120);draw(O -- (O+E) -- (O+E+4*NE) -- (O+E+4*NE+2*NW) -- (O-3*E+4*...

We see that the figure contains 43 unit triangles, and therefore its area is \boxed{\frac{43\sqrt{3}}4}}.

See also

1999 AHSME (Problems)
Preceded by
Problem 24
Followed by
Problem 26
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