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# Geometry Question

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$$ABCD$$ is an isosceles trapezoid. The bases of this trapezoid are  $$\overline{AB}$$ and $$\overline{CD}$$. A circle is inscribed in the quadrilateral (the circle is tangent to all the sides of the trapezoid.) Base $$\overline{AB}$$ has a length of $$2x$$ and base $$\overline{CD}$$ has a length of $$2y$$. Prove that the radius of the inscribed circle is $$\sqrt{xy}$$.

Diagram: Jun 8, 2021

#1
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Try using Brahmagupta's theorem.

Jun 8, 2021
#2
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Jun 9, 2021
#3
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thanks for all the help i figured out the answer

Jun 9, 2021
#4
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First, let's label all the things we know on the trapezoid. Label AB as 2x and CD as 2y.

Now, draw the center of the circle and label it as O. Draw the diameter of the circle that is perpendicular to AB and CD. Then draw a line from O to B and C Next draw a line from O to BC that is perpendicular and tangent to BC and make the point E.

The diameter of the circle intercepts with AB and cuts it in half. Let's call the point F. Because it cuts AB in half, FB is (2x)/2=x.

The diameter of the circle also intercepts with CD and cuts it in half. Let's call the point G. Because it cuts CD in half, GC is (2y)/2=y.

Using the property of a circle we know that FB and EB are the same length, making them both x. The same thing goes with FB and EB which are y.

Now draw a line from B to CD that is perpendicular to CD and label the point H.

BH is 2r because it is the same length as the diameter of the circle. BC is x+y, and HC is y-x.

Using the pythagorean theorem, we can say that (2r)^2+(y-x)^2=(x+y)^2.
Now, let's solve the equation.

(2r)^2+(y-x)^2=(x+y)^2
4r^2+(y-x)^2=(x+y)^2
4r^2=(x+y)^2-(y-x)^2
4r^2=x^2+2xy+y^2-(y^2-2xy+x^2)
4r^2=x^2+2xy+y^2-y^2+2xy-x^2
4r^2=x^2-x^2+2xy+2xy+y^2-y^2
4r^2=4xy
r^2=xy
r=xy.

Jun 9, 2021