Geometry Problem

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 908.

## Monday, August 5, 2013

### Problem 908: Bicentric Quadrilateral, Incircle, Circumcircle, Circunscribed, Inscribed, Tangent, Inradius

Labels:
bicentric quadrilateral,
incenter,
incircle,
inradius

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1/(BI)^2 + 1/(DI)^2 = 1/(AI)^2 + 1/(CI)^2 = 1/x^2 -- Property Bicentric Quad. or 1/16 + 1/4 =1/x^2 or x = 4/√5

ReplyDeleteCG=sqrt(4-x^2)

ReplyDeleteICG~AIH

sqrt(4-x^2)/x=0.5

1.25x^2=4

x=2/sqrt(1.25)