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