Friday, March 13, 2015

Geometry Problem 1097: Quadrilateral, Inscribed Circle, 90 Degree, Angle, Area

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.

Online Math: Geometry Problem 1097: Quadrilateral, Inscribed Circle, 90 Degree, Angle, Area.

4 comments:

  1. Let AB=a, BC=b, CDgc, DA=d.

    Then
    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)

    ReplyDelete
  2. Area ABCD=2 Area ABC(deoarece A,OsiC sunt coliniare)=2(AreaAOB+AreaBOC)=2(AB.r/2+BC.r/2)=r(AB+BC)

    ReplyDelete
  3. We can see there are two squares OT_1BT_2 and OT_3DT_4. That means AB=AD and BC=BD.
    (ABCD)=(ABO)+(BOC)+(COD)+(DOA)=1/2 r( AB+BC+CD+DA)=R(AB+BC)

    ReplyDelete
  4. It also follows that r^2,= AT1. CT2

    ReplyDelete