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

Subscribe to:
Post Comments (Atom)

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

ReplyDelete