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Thread: Problem finding area

  1. #1
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    Problem finding area

    i am having a problem finding the area of the three triangles outside the Pitagoras drawing
    you can see what i am meaning in this picture
    http://i51.tinypic.com/28gxmyd.jpg
    so i need the area of all the triangles in the middle of the squares

    Thanks in advance
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  2. #2
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    Re: Problem finding area

    Hello, leart369!

    http://i51.tinypic.com/28gxmyd.jpg
    i need the area of all the triangles.

    Label the inner right triangle like this:

    Code:
          A
          *
          |   *     c
        b |       *
          |           *
          * - - - - - - - * B
        C         a

    Note that for any triangle, its area is: .$\displaystyle A \;=\;\tfrac{1}{2}ab\sin C$
    . . (One-half the product of two sides and the sine of the included angle)

    The right triangle at the lower-left has sides $\displaystyle a$ and $\displaystyle b$.
    . . Its area is: .$\displaystyle \boxed{A_1 \:=\:\tfrac{1}{2}ab}$

    The triangle at the right has sides $\displaystyle a$ and $\displaystyle c$, and included angle $\displaystyle (180^o - B).$
    . . Its area is: .$\displaystyle A_2 \:=\:\tfrac{1}{2}ac\sin(180^o-B) \:=\:\tfrac{1}{2}ac\sin B$
    In the above diagram, we see that: .$\displaystyle \sin B \,=\,\tfrac{b}{c}$
    . . Hence: .$\displaystyle A_2 \;=\;\tfrac{1}{2}ac\left(\tfrac{b}{c}\right) \quad\Rightarrow\quad \boxed{A_2 \:=\:\tfrac{1}{2}ab}$

    The third triangle has sides $\displaystyle b$ and $\displaystyle c$, and included angle $\displaystyle (180^o - A).$
    . . Its area is: .$\displaystyle A_3 \:=\:\tfrac{1}{2}bc\sin(180^o-A) \:=\:\tfrac{1}{2}bc\sin A$
    In the above diagram, we see that: .$\displaystyle \sin A \,=\,\tfrac{a}{c}$
    . . Hence: .$\displaystyle A_3 \;=\;\tfrac{1}{2}bc\left(\tfrac{a}{c}\right) \quad\Rightarrow\quad \boxed{A_3 \:=\:\tfrac{1}{2}ab}$


    Fascinating! . . . All four triangles have the same area.

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  3. #3
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    Re: Problem finding area

    does exist any way to find the area off all that drawing?
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  4. #4
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    Re: Problem finding area

    Quote Originally Posted by leart369 View Post
    does exist any way to find the area off all that drawing?
    1. I assume that you mean the area of the hexagon, containing 3 squares, 2 right triangles and 2 obtuse triangles. If so:

    2. As Soroban has shown you have:

    $\displaystyle \text{complete area} = a^2+b^2+c^2+4 \cdot \frac12 \cdot a \cdot b = c^2 +(a+b)^2$

    3. Since $\displaystyle a^2+b^2 = c^2$ there are other possibilities to simplify the sum of areas.
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  5. #5
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    Re: Problem finding area

    thank you very much that was what i was looking for
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