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Math Help - Find Area of Right Triangle

  1. #1
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    Find Area of Right Triangle

    Right triangle, where one of the legs equals 10, the bisector of the opposite angle cuts the leg into the ratio of 2:3. Find the area of the triangle.
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  2. #2
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    Hello, magentarita!

    In a right triangle, where one of the legs equals 10,
    the bisector of the opposite angle cuts the leg into the ratio of 2:3.
    Find the area of the triangle.

    This problem requires a seldom-used theorem about angle bisectors.

    In a triangle, the angle bisector of an angle divides the opposite side
    . . into segments proportional to the other two sides.

    Code:
                  B
                  *
                 *  *  2a
                *     *
             2 *        *   D
              *           *
             *        *     *  3a
            *     *           *
           *  *                 *
        A *   *   *   *   *   *   *  C
                        3
    Suppose we have \Delta ABC with AB = 2,\:AC = 3

    If AD bisects \angle A, then BD:DC \,=\,2:3




    We have the following right triangle . . .
    Code:
                              B
                              o   -
                            * *   |
                          *   *   |
                        *   3 *   |
                  3a  *       *   10
                    *         o D |
                  *       *   *   |
                *     *       *   |
              *   *         2 *   |
            * *               *   |
        A o   *   *   *   *   o   -
                    2a        C

    We are given: . BC = 10
    AD bisects \angle A and divides BC in the ratio 2:3.

    Hence, AC:AB \,=\,2:3

    Let: AC = 2a,\;AB = 3a

    Using Pythagorus, we have: . (2a)^2 + 10^2 \:=\:(3a)^2

    Then: . 4a^2 + 100 \:=\:9a^2\quad\Rightarrow\quad 5a^2 \:=\:100 \quad\Rightarrow\quad a^2 \:=\:20

    . . Hence: . a \:=\:\sqrt{20} \:=\:2\sqrt{5}

    So we have: . AC \:=\:2a \:=\:4\sqrt{5}


    The area of the triangle is: . \tfrac{1}{2}\text{(base)(height)} \:=\: \tfrac{1}{2}(4\sqrt{5})(10) \;=\; 20\sqrt{5}

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  3. #3
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    Just out of curiousity , is there a name for this theorem and is there a proof for this theorem . THanks .
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  4. #4
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    Soroban...

    Quote Originally Posted by Soroban View Post
    Hello, magentarita!


    This problem requires a seldom-used theorem about angle bisectors.

    In a triangle, the angle bisector of an angle divides the opposite side
    . . into segments proportional to the other two sides.
    Code:
                  B
                  *
                 *  *  2a
                *     *
             2 *        *   D
              *           *
             *        *     *  3a
            *     *           *
           *  *                 *
        A *   *   *   *   *   *   *  C
                        3
    Suppose we have \Delta ABC with AB = 2,\:AC = 3

    If AD bisects \angle A, then C \,=\,2:3" alt="BDC \,=\,2:3" />




    We have the following right triangle . . .
    Code:
                              B
                              o   -
                            * *   |
                          *   *   |
                        *   3 *   |
                  3a  *       *   10
                    *         o D |
                  *       *   *   |
                *     *       *   |
              *   *         2 *   |
            * *               *   |
        A o   *   *   *   *   o   -
                    2a        C
    We are given: . BC = 10
    AD bisects \angle A and divides BC in the ratio 2:3.

    Hence, AC:AB \,=\,2:3

    Let: AC = 2a,\;AB = 3a

    Using Pythagorus, we have: . 3a)^2" alt="(2a)^2 + 10^2 \:=\3a)^2" />

    Then: . 4a^2 + 100 \:=\:9a^2\quad\Rightarrow\quad 5a^2 \:=\:100 \quad\Rightarrow\quad a^2 \:=\:20

    . . Hence: . a \:=\:\sqrt{20} \:=\:2\sqrt{5}

    So we have: . AC \:=\:2a \:=\:4\sqrt{5}


    The area of the triangle is: . \tfrac{1}{2}\text{(base)(height)} \:=\: \tfrac{1}{2}(4\sqrt{5})(10) \;=\; 20\sqrt{5}
    Wonderful job as usual. Thanks.
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