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2005 Alabama ARML TST Problems/Problem 11

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Problem

In concave hexagon $ABCDEF$ , $\angle A = \angle B = \angle C = 90^\circ$ , $\angle D = 100^\circ$ , and $\angle F = 80^\circ$ . Also, $CD=FA$ , $AB=7$ , $BC=9$ , and $EF+DE=12$ . Compute the area of the hexagon.

Solution

Join four such hexagons and let the points be labeled as shown below:

Since the sum of the angles in a hexagon is $(6-2)180^\circ = 720^\circ$ , $\angle E = 720^\circ - (3\cdot 90^\circ + 80^\circ + 100^\circ) = 270^\circ$ .

Since $\angle D_iE_iF_i = 360^\circ - \angle E = 360^\circ - 270^\circ = 90^\circ$ and $\angle E_iF_iE_{i+1} = \angle F + \angle D = 80^\circ + 100^\circ = 180^\circ$ , $E_1E_2E_3E_4$ is a square with a side length of $EF+DE = 12$ .

Also, Since $\angle A_iB_iC_i = \angle B = 90^\circ$ and $\angle B_iA_iB_{i+1} = \angle A + \angle C = 90^\circ + 90^\circ = 180^\circ$ , $B_1B_2B_3B_4$ is a square with a side length of $AB+BC = 7+9 = 16$ .

Therefore, $4[ABCDEF] + [E_1E_2E_3E_4] = [B_1B_2B_3B_4] \Longrightarrow$ $[ABCDEF] = \frac{[B_1B_2B_3B_4] - [E_1E_2E_3E_4]}{4} = \frac{16^2-12^2}{4} = \boxed{28}$ .

See also

2005 Alabama ARML TST (Problems)
Preceded by: Problem 10	Followed by: Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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