Tuesday, June 7, 2011

Problem 615: Right Triangle with a Square, Diagonals, Center, Area of Quadrilateral

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

Click the figure below to see the complete problem 615.

Problem 615: Right Triangle with a Square, Diagonals, Center, Area of Quadrilateral

4 comments:

  1. OABC is a cyclic quadrilateral
    By 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

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  2. http://s30.postimg.org/qiktgwn6p/pro_615.png

    Solution of this problem depend on value b , length of side of square ACDE.
    Case 1: b< 6, we have no solution
    Case 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.

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  3. Area • ac/2 + b^2/4 = (a+c)^2 / 4 using Pythagoras

    Applying Ptolemy to cyclic quad ABCO

    ab/sqrt2 + cb/sqrt2 = 6b

    Hence a+c = 6aqrt2 and so area = 36 X 2 /4 = 18

    Sumith Peiris
    Moratuwa
    Sri Lanka

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