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Area of Pentagon (Posted on 2008-01-07) Difficulty: 3 of 5
  ABCDE is a pentagon with
   |AB| = |BC|, |CD| = |DE|,

   /ABC = 150°, /CDE = 30°,

   and |BD| = 2.
Find the area of ABCDE.
 

See The Solution Submitted by Bractals    
Rating: 2.0000 (1 votes)

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Solution two ways (spoiler) | Comment 1 of 4

Playing around with Geometer's Sketchpad, the area's relation to BD^2 remains a constant at what appears to be |BD|^2 / 4, which would come out to 4/4 = 1.

Analytically, if we assume this puzzle has one answer, regardless of the specific configuration of those features that can be varied, then if we let point E approach point A, the area will approach that of what then becomes quadrilateral ABCD. Since there is considered to be only one answer, this must be a constant, and so the area that is approached must be the answer.

Since AB=BC and CD=DA (when pt A = pt E), the figure is a kite. Angles C and A are equal and add up to 180 degrees and so are right angles. The kite consists of two right triangles joined at their common hypotenuse BD.

The area of this kite is |AB||BD|.

The angles of the right triangle are half those of the kite, and so are 15 degrees and 75 degrees.

Using the half-angle formulas:

sin 15 = sqrt((1-sqrt(3)/2)/2)
cos 15 = sqrt((1+sqrt(3)/2)/2)

The area sought is |AB||BD|.= 2 sin 15 * 2 cos 15 = 4 sin 15 cos 15.

That is, 4 sqrt((1 - 3/4) / 4) = 1


  Posted by Charlie on 2008-01-07 13:45:11
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