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Let ABCD be an isosceles trapezoid, with bases AB and CD. and  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 sqrt(xy)

 

https://gyazo.com/4dd0a1236ad5c782bc9d82015f5cba3b

 Mar 31, 2020
 #1
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Image is blocked on my browser! Can you provide the picture in your question? Thanks

 Mar 31, 2020
 #2
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Not sure how to upload image, just a trapezoid with top base AB and lower base DC respectively, with a circle inscribed in it so the circle is tangent to every side. 

Guest Mar 31, 2020
 #3
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Working on it right now.

 #5
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Let ABCD be an isosceles trapezoid, with bases AB and CD. and  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 sqrt(xy).

 

I'll use the real numbers:

 

AB = 2x = 5    x = 2.5         CD = 2y = 10     y = 5

AD = BC = AB/2 + CD/2 = 7.5

Radius r = ?          r =  {sqrt [(AD)² - (CD/4)²]} /2  = 3.535533906

 

                                       sqrt (xy) = sqrt (2.5 * 5) = 3.535533906

 

Dragan  Apr 1, 2020
edited by Dragan  Apr 1, 2020
 #6
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An alternate method, I always like those!

 #4
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Hello Guest! Make an account!

Set up:

- By the two tangent theorem, we can find the length of the sides of the trapezoid.

- I calculated the shorter leg of the right triangle with hypotenuse AD by doing (DC - AB) / 2

- The height of the longer leg of the right triangles is equal to the diameter indicated by the red line.

Calculate:

By pythagorean theorem

(x + y)2 - (x - y)2 =

x2 + 2xy + y2 - (x2 -2xy + y2) = 

4xy

 

Diameter = sqrt(4xy)

Diameter = \(2\sqrt{xy}\)

Radius = \(\sqrt{xy}\)

 

cheeky

 Apr 1, 2020

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