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Tangent Trapezoid (Posted on 2010-07-28) Difficulty: 3 of 5

Let (P,r) denote a circle with center P and radius r.

Circles (A,a) and (B,b) are externally tangent with 
a less than b.

Let CD and EF be the common external tangents to
circles (A,a) and (B,b) with C and F on circle (A,a) 
and D and E on circle (B,b).

What is the area of trapezoid CDEF in terms of 
a and b?

  Submitted by Bractals    
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Solution: (Hide)

Let [KLM...XYZ] denote the area of polygon KLM...XYZ.

Let the common internal tangent intersect CD and EF
at points P and R respectively. Let Q be the intersection
of AB and PR. Tangents to a circle from an external 
point have the same length. Therefore,

   |PC| = |PQ| = |PD|

        and 

   |RE| = |RQ| = |RF|

By symmetry, |PQ| = |RQ|. Therefore,

   |CD| = |PR| = |EF|

Construct PS perpendicular to EF with S on EF and
AG perpendicular to BE with G on BE. 

Triangles PSR and AGB are similar since corresponding
sides are perpendicular to each other. Therefore,

    |PS|     |AG|
   ------ = ------.
    |PR|     |AB|

Since |AG| = |EF|, we have

   |EF|2 = |AB|2 - |BG|2 = (b + a)2 - (b - a)2 = 4ab

         and

   |PS| = |EF|2/|AB| = 4ab/(a + b)

See first

    |CP|     |ER|
   ------ = ------, implies
    |PD|     |RF| 

   [CDEF] = [PEF] + [RCD] = 2[PEF]

          = |PS||EF| = 8ab*sqrt(ab)/(a + b).


-----------------------------------------------------------

See Daniel's post for an alternate solution.


Comments: ( You must be logged in to post comments.)
  Subject Author Date
analytical solutionDaniel2010-07-28 16:26:33
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