Geometry Problem

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## Friday, February 11, 2011

### Problem 578: Circle, Perpendicular Chords, Midpoints, Curvilinear Triangle, Area, 90 Degrees

Labels:
90,
area,
chord,
circle,
curvilinear triangle,
degree,
midpoint,
perpendicular

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draw diameter MN perpendicular to AB,

ReplyDeletediameter KL perpendicular to CD

A1 - CGOM -2A5 - OFBL = A3 - AKGE + CGOM

A1 - A3 = 2A5 + 2CGOM + OFBL - AKGE (1)

AKGE + 2A5 = OFBL (2)

A2 + EFND = A4 - EFND

A2 - A4 = - 2EFND (3)

A1 + A2 - A3 - A4 = 2A5 + (OFBL - AKGE)+ (2CGOM -2EFND)

A1 + A2 - A3 - A4 = 2A5 + 2A5 + 4A5

A1 + A2 - A3 - A4 = 8A5

http://img832.imageshack.us/img832/1769/problem578.png

ReplyDeleteLet O is the center of the circle.

Draw segments HFG and ML symmetric of CED and EF over center O ( see picture)

By observation we have A1-A3= A(EFHC) and A2-A4= -A(EFGD)

But A(EFGD)=A(CMLH) ( symmetric property)

So A1-A3+A2-A4 = A(EFHC)-A(CMLH)=A(MEFL)= 8x A5

Peter Tran