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?

Guest Aug 15, 2023

#1**+1 **

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

CPhill Aug 15, 2023