Difference between revisions of "2005 Alabama ARML TST Problems/Problem 11"
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==Solution== | ==Solution== | ||
− | {{ | + | Join four such hexagons and let the points be labeled as shown below: |
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+ | [[Image:Fourhexsquare.JPG]] | ||
+ | |||
+ | Since the sum of the angles in a hexagon is <math>(6-2)180^\circ = 720^\circ</math>, <math>\angle E = 720^\circ - (3\cdot 90^\circ + 80^\circ + 100^\circ) = 270^\circ</math>. | ||
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+ | Since <math>\angle D_iE_iF_i = 360^\circ - \angle E = 360^\circ - 270^\circ = 90^\circ</math> and <math>\angle E_iF_iE_{i+1} = \angle F + \angle D = 80^\circ + 100^\circ = 180^\circ</math>, <math>E_1E_2E_3E_4</math> is a square with a side length of <math>EF+DE = 12</math>. | ||
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+ | Also, Since <math>\angle A_iB_iC_i = \angle B = 90^\circ</math> and <math>\angle B_iA_iB_{i+1} = \angle A + \angle C = 90^\circ + 90^\circ = 180^\circ</math>, <math>B_1B_2B_3B_4</math> is a square with a side length of <math>AB+BC = 7+9 = 16</math>. | ||
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+ | Therefore, <math>4[ABCDEF] + [E_1E_2E_3E_4] = [B_1B_2B_3B_4] \Longrightarrow</math> <math>[ABCDEF] = \frac{[B_1B_2B_3B_4] - [E_1E_2E_3E_4]}{4} = \frac{16^2-12^2}{4} = \boxed{28}</math>. | ||
==See also== | ==See also== | ||
{{ARML box|year=2005|state=Alabama|num-b=10|num-a=12}} | {{ARML box|year=2005|state=Alabama|num-b=10|num-a=12}} |
Latest revision as of 23:31, 16 June 2008
Problem
In concave hexagon , , , and . Also, , , , and . 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 , .
Since and , is a square with a side length of .
Also, Since and , is a square with a side length of .
Therefore, .
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 |