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Thread: Area of a triangle Question

  1. #1
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    Area of a triangle Question

    Hi,

    I found this question and I could do with some help regarding the answer. It is:

    Nine lines are drawn parallel to the base of a triangle to divide each of the other sides into 10 equal segments. If the area of the largest region is 38, find the area of the original triangle.

    I believe that it must be ratios and something like:
    Area of a triangle Question-untitled.jpg
    (Sorry bad drawing!)

    Ratio of sides = $\displaystyle 10x:9x$
    Therefore, ratio of areas = $\displaystyle 100x^2 : 81x^2$
    $\displaystyle 100x^2-81x^2=38$
    $\displaystyle 19x^2=38$
    $\displaystyle x^2=2$

    Therefore area = $\displaystyle 100x^2=200$

    Am I right? I feel that this is rather inadequate and too simple.

    Thanks very much for any help!
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  2. #2
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    Re: Area of a triangle Question

    To "see" how this works, start with a right triangle: 10 by 24 by 26 will make it easier...
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  3. #3
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    Re: Area of a triangle Question

    Alternatively you could use the standard formula for the area of a triangle with sides $\displaystyle a,b$ and included angle $\displaystyle \theta.$

    Area = $\displaystyle \frac{1}{2}ab\sin\theta.$

    Call the sloping sides $\displaystyle 10m$ and $\displaystyle 10n$ say, then the area of the last strip will be

    $\displaystyle (\frac{1}{2})100mn\sin\theta - (\frac{1}{2})81mn\sin\theta = (\frac{1}{2})19mn\sin\theta = 36,$ in which case $\displaystyle mn\sin\theta=4,$ so that the area of the whole triangle will be
    $\displaystyle (\frac{1}{2}).100.4 = 200.$
    Thanks from BobtheBob
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  4. #4
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    Re: Area of a triangle Question

    A model triangle has an altitude of 10 units and a base of10 units. Its area = 50 sq units
    Area of largest trapezoid section = 50 - 9*9/2 =9.5 sq units
    Fractional area of above 9.5/50 = 0.19
    Area of actual triangle* 0.19 =38
    AAT = 200 sq units
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  5. #5
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    Re: Area of a triangle Question

    Hello, BobtheBob!

    Here's another approach . . .


    Nine lines are drawn parallel to the base of a triangle to divide each of the other sides into 10 equal segments.
    If the area of the largest region is 38, find the area of the original triangle.

    Code:
        -         *
        :        *  *
        :       *     *
        :      *        *
        h     *           *
        :    *   (9/10)b    *
        :   * * * * * * * * * * - - -
        :  *                    *  (1/10)h
        - *  *  *  *  *  *  *  *  * -
          : - - - - - b - - - - - :
    The base of the triangle is $\displaystyle b$; its height is $\displaystyle h.$

    The bottom trapezoid has height $\displaystyle \tfrac{1}{10}h$ and bases $\displaystyle b$ and $\displaystyle \tfrac{9}{10}b.$

    Its area is 38: .$\displaystyle \tfrac{1}{2}\left(\tfrac{1}{10}h\right)\left(b + \tfrac{9}{10}b\right) \:=\:38 \quad\Rightarrow\quad \tfrac{19}{200}bh \:=\:38 \quad\Rightarrow\quad bh \:=\:400$

    Therefore: .$\displaystyle \text{Area}_{\Delta} \:=\:\tfrac{1}{2}bh \:=\:200$
    Thanks from BobtheBob
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  6. #6
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    Thumbs up Re: Area of a triangle Question

    Thanks! So I was right with 200. However, all the solutions here are far better than mine in my opinion!

    Thanks again for the help!
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  7. #7
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    Re: Area of a triangle Question

    Quote Originally Posted by BobtheBob View Post
    Nine lines are drawn parallel to the base of a triangle to divide each of the other sides into
    10 equal segments. If area of the largest region is 38, find the area of the original triangle.
    Givens are:
    j = area largest region (38)
    h = height of triangle (10)
    n = number of regions (10)
    Find:
    k = area of triangle (?)

    k = bh / 2
    where b = 2jn(n - 2) / [2(h/n)(n^2 - 3n + 1) + h]
    (b = base of triangle)

    Substitute the givens and you'll get b = 40, hence k = 40(10) / 2 = 200

    Now Bob, you can use that as a "general case" formula;
    weird ones like j = 137, h = 23, n = 7 handled quickly!
    That one will give you b = 44.903.... and k = 516.384....
    Thanks from BobtheBob
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