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Geometry ProblemLevel: Mathematics Education, High School, College.Click the figure below to see the complete problem 615.
OABC is a cyclic quadrilateralBy Ptolemy's Theorem:AB.OC+BC.OA=OB.AC(c+a)(b/√2)=6b, c+a=6√2Required area =(ABC)+(COA) =(1/2)ca + (b²/4)=(1/4)(2ca+c²+a²)by Pythagoras=(1/4)(c + a)²=18 sq units
Problem 615: This solution was submitted by Michael Tsourakakis from Greece Thanks Michael
http://s30.postimg.org/qiktgwn6p/pro_615.pngSolution of this problem depend on value b , length of side of square ACDE. Case 1: b< 6, we have no solutionCase 2: 6< b< 6.sqrt(2) we have 2 solutions. Area of ABCO is 18 Case 3: b> 6.sqrt(2) , we have 2 solutions. Point B is inside the square ACDE And the area of ACBO is undefined.
Area • ac/2 + b^2/4 = (a+c)^2 / 4 using Pythagoras Applying Ptolemy to cyclic quad ABCOab/sqrt2 + cb/sqrt2 = 6bHence a+c = 6aqrt2 and so area = 36 X 2 /4 = 18Sumith PeirisMoratuwaSri Lanka