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Quadrilateral ABCD is an isosceles trapezoid, with bases AB and CD. A circle is inscribed in the trapezoid, as shown below. (In other words, the circle is tangent to all the sides of the trapezoid.) The length of base AB is 2x and the length of base CD is 2y. Prove that the radius of the inscribed circle is √xy .

 Jan 28, 2021
 #1
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If x = y, then the quadrilateral is a square, and r = sqrt(xy) = x.  So it's true!

 Jan 28, 2021
 #2
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Drop a vertical from A meeting DC at E.

DE = (2y - 2x)/2 = y-x.

AD = x + y.

AE = 2R.

Now Pythagoras, AD^2 = AE^2 + DE^2, etc..

 Jan 28, 2021
 #3
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Drop  perpendicular  BE   from B to  DC

 

Then EC  =  (2y  -2x)  /2   =  y  - x

 

And  by a property of  tangents drawn to a circle,  BC  = x +  y 

 

So

 

BE^2    =   BC^2  -  EC^2

 

BE^2  =  ( x^2 + 2xy + y^2)  - ( x^2  - 2xy + y^2)

 

BE^2   =  4xy

 

BE   = 2sqrt (xy)

 

But the radius is  (1/2)  of BE  =  sqrt (xy)

 

 

cool cool cool

 Jan 28, 2021

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