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

7 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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  4. Append triangles congruent to ABC to the other sides of the square, thus creating a new square called BFGH. Connect BO up through G. Since 0 is the center of both squares, then BO is 12. The area of the BFGH is 72 = 12^2 / 2. Since quadrilateral AOCB is 1/4 of the area of the square BFGH, then the area of AOCB is 18.

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  5. https://photos.app.goo.gl/UGeJgNArp2tU6dzJ8

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  6. Let the length of the square be x
    <AOC=90=<ABC
    ABCD is a cyclic quad
    By Ptolemy Theorem, (AO)(BC)+(AB)(CO)=(AC)(BO)
    (x/sqrt2)*(BC)+(x/sqrt2)*(AB)=6x
    AB+BC=6sqrt2
    AB^2+BC^2=x^2
    2(AB)(BC)=72-x^2
    (AB)(BC)/2=18-x^2/4
    (AO)(OC)/2=x^2/4
    [OABC]=[ABC]+[AOC]=[(AB)(BC)/2]*[(AO)(OC)/2]=18

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