Geometry Problem
Level: Mathematics Education, High School, College.
Click the figure below to see the complete problem 615.
Tuesday, June 7, 2011
Problem 615: Right Triangle with a Square, Diagonals, Center, Area of Quadrilateral
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OABC is a cyclic quadrilateral
ReplyDeleteBy Ptolemy's Theorem:
AB.OC+BC.OA=OB.AC
(c+a)(b/√2)=6b, c+a=6√2
Required area
=(ABC)+(COA)
=(1/2)ca + (b²/4)
=(1/4)(2ca+c²+a²)by Pythagoras
=(1/4)(c + a)²
=18 sq units