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Math Help - Rhombus Problem

  1. #1
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    Rhombus Problem

    Given Rhombus ABCD (not shown) AB = 10 and AC = 12. Find AD and BD

    I know that AD = 10 because the sides of a rhombus are all congruent. I cannot find what BD equals though. I thought it was 12 but I don't think the diagonals of a rhombus are congruent.
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    Re: Rhombus Problem

    Quote Originally Posted by Cake View Post
    Given Rhombus ABCD (not shown) AB = 10 and AC = 12. Find AD and BD

    I know that AD = 10 because the sides of a rhombus are all congruent. I cannot find what BD equals though. I thought it was 12 but I don't think the diagonals of a rhombus are congruent.
    Rhombus Problem-clipboard01.jpg

    x^2+y^2=10^2=100

    (10-x)^2+y^2=12^2=144

    (10+x)^2+y^2=BD^2

    solving the first two equations we get x=\frac{14}{5}$ and $y=\frac{48}{5}

    plugging into the 3rd equation we get

    \left(10+\frac{14}{5} \right)^2+\left(\frac{48}{5} \right)^2=BD^2

    \left(\frac{64}{5}\right)^2+\left(\frac{48}{5} \right)^2=\frac{6400}{25}=BD^2\Rightarrow BD=\frac{80}{5}=16
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    Re: Rhombus Problem

    Quote Originally Posted by romsek View Post
    Click image for larger version. 

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    x^2+y^2=10^2=100

    (10-x)^2+y^2=12^2=144

    (10+x)^2+y^2=BD^2

    solving the first two equations we get x=\frac{14}{5}$ and $y=\frac{48}{5}

    plugging into the 3rd equation we get

    \left(10+\frac{14}{5} \right)^2+\left(\frac{48}{5} \right)^2=BD^2

    \left(\frac{64}{5}\right)^2+\left(\frac{48}{5} \right)^2=\frac{6400}{25}=BD^2\Rightarrow BD=\frac{80}{5}=16
    Holy cow! That's some math work! Thank you, is there any shorter way than this?
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    Re: Rhombus Problem

    Quote Originally Posted by Cake View Post
    Holy cow! That's some math work! Thank you, is there any shorter way than this?
    Yes. The diagonals of a rhombus bisect each other at right angles.AB =10 AC =12. Note that there are four congruent triangles formed by them.1/2 of AC =6. 1/2 of BD=8 5-4- 3 right triangle
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    Re: Rhombus Problem

    Quote Originally Posted by bjhopper View Post
    Yes. The diagonals of a rhombus bisect each other at right angles.AB =10 AC =12. Note that there are four congruent triangles formed by them.1/2 of AC =6. 1/2 of BD=8 5-4- 3 right triangle
    Right, so half of AC is 6. For example, if we had point E in the middle of the rhombus. AE would be 6 because it is half of 12. then AB is 10. You just don't add 10 and 6 to get 16, right?
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    Re: Rhombus Problem

    Quote Originally Posted by Cake View Post
    Right, so half of AC is 6. For example, if we had point E in the middle of the rhombus. AE would be 6 because it is half of 12. then AB is 10. You just don't add 10 and 6 to get 16, right?
    hrm I should have seen this.

    No what you do is notice that

    \left(\frac{BD}{2}\right)^2+\left(\frac{12}{2} \right)^2=10^2

    \left(\frac{BD}{2}\right)^2=100-36=64 \Rightarrow \frac{BD}{2}=8 \Rightarrow BD=16
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    Re: Rhombus Problem

    Hello, Cake!

    \text{Given rhombus }ABCD,\;AB = 10\text{ and }AC = 12.\;\text{ Find }AD\text{ and }BD.

    I know that AD = 10 because the sides of a rhombus are all congruent.
    I cannot find what BD equals though.
    I thought it was 12, but I don't think the diagonals of a rhombus are congruent.
    If they were, you'd have a square.

    The diagonals of a rhombus are perpendicular and bisect each other.

    Hence: . AO = OC = 6.

    Code:
                 A       10        B
                  o---------------o
                 / *           * /
                /   *6      *   /
               /     *   *     /
           10 /       o       /
             /     *  O*     /
            /   *       *6  /
           / *           * /
          o---------------o
         D                 C
    In right triangle AOB, we find that OB = 8.

    Therefore: . BD = 16.
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    Re: Rhombus Problem

    Quote Originally Posted by Soroban View Post
    Hello, Cake!


    The diagonals of a rhombus are perpendicular and bisect each other.

    Hence: . AO = OC = 6.

    Code:
                 A       10        B
                  o---------------o
                 / *           * /
                /   *6      *   /
               /     *   *     /
           10 /       o       /
             /     *  O*     /
            /   *       *6  /
           / *           * /
          o---------------o
         D                 C
    In right triangle AOB, we find that OB = 8.

    Therefore: . BD = 16.

    Makes sense ^_____^
    Thank you guys!
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