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Two circles, centered at A and B have radius 1 and 2, respectively, with AB=5. The common internal tangent intersects the smaller circle at C and the larger circle at D what is the area of quadrilateral ACBD?

 

 Aug 15, 2023
 #1
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See the following

 

 

Angle BCE  = 90    = Angle ADE

Because BC and DA are parallel, then angles EAD and EBC are opposite interior angles between these  parallels.....therefore they are equal

 

So by AA congruency, triangles EAD and EBC are similar

 

And  AD = 2 and BC = 1......therefore the ratio of the  sides = 2 : 1

 

Then BE =  [ 1 / (1 + 2) ] (AB)  =  ( 1/3)(5)  = 5/3

And by the Pythagoren Theorem, CE = sqrt ( BE^2 - BC^2) =  sqrt (5/3^2 - 1^2)  = sqrt (16/ 9) = 4/3

And AE =  5 - 5/3 =  10/3

And by the P Theorem again , ED = sqrt (  AE^2 - DA^2) =  sqrt [ (10/3)^2 - 2^2] =  sqrt [ 64/9] = 8/3

 

So CD =  CE + ED =  4/3 + 8/3 =  12/3 =  4

 

And ACBD is a parallelogram  with bases of 1,2   and  a height of 4

 

So [ ACBD ]  =  (1/2) (height) ( sum of the bases)  = (1/2) (4) ( 1 +2)  = 6 

 

 

cool cool cool

 Aug 15, 2023

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