Geometry Problem. Post your solution in the comment box below.

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

Problem submitted by Maurice Ho

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## 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

ReplyDelete[ABCD]=(r/2)*(AB+BC+CD+DA)

ReplyDelete=(r/2)*[AT1+BT1+BT2+CT2+CT3+DT3+DT4+AT4]

=(r/2)[AT1+BT1+BT2+CT2+CT2+DT4+DT4+AT1]

=(r/2)[2AT1+2CT2+2DT4+BT1+BT2]-------(1)

DT4=OT3=r=OT1=OT2=BT1=BT2

2DT4=B1+B2------(2)

Sub (2) in (1)

[ABCD]=(r/2)[2A1+2CT2+2BT1+2BT2]=(r/2)[2(AB+BC)]=r(AB+BC)

T1BT2O & T3OT4D are squares with side r

ReplyDeleteSo AB = AD and BC = CD and so S[ABC] = S[ACD]

Hence S[ABCD] = 2 X S[ABC] = 2.(1/2 r.AB + 1/2 rBC) = r.(AB + BC)

Sumith Peiris

Moratuwa

Sri Lanka