Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
Problem submitted by Maurice Ho
Click the diagram below to enlarge it.
Friday, March 13, 2015
Geometry Problem 1097: Quadrilateral, Inscribed Circle, 90 Degree, Angle, Area
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Let AB=a, BC=b, CDgc, DA=d.
ReplyDeleteThen
a²+b² = c²+d²
a+c = b+d
a-b = d-c
square then gives ab=cd
Thus a+b=d+c
Hence, a=d and b=c.
Area = 1/2 r(a+b+c+d) = r(a+b)
Area ABCD=2 Area ABC(deoarece A,OsiC sunt coliniare)=2(AreaAOB+AreaBOC)=2(AB.r/2+BC.r/2)=r(AB+BC)
ReplyDeleteWe can see there are two squares OT_1BT_2 and OT_3DT_4. That means AB=AD and BC=BD.
ReplyDelete(ABCD)=(ABO)+(BOC)+(COD)+(DOA)=1/2 r( AB+BC+CD+DA)=R(AB+BC)
It also follows that r^2,= AT1. CT2
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