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Thread: Help request on the pythagorean Theorem 2.

  1. #1
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    Help request on the pythagorean Theorem 2.

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  2. #2
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    $\displaystyle OC^2 + AC^2 = OA^2$

    But $\displaystyle OC = AC$. I'll let both equal h:
    $\displaystyle \begin{aligned}
    OC^2 + AC^2 &= r^2 \\
    h^2 + h^2 &= r^2 \\
    2h^2 &= r^2 \\
    h^2 &= \frac{r^2}{2} \\
    h &= \frac{r\sqrt{2}}{2}
    \end{aligned}$

    Area of a triangle is $\displaystyle \frac{1}{2}bh$. The base is OB and the height is AC:
    $\displaystyle \begin{aligned}
    \frac{1}{2}bh &= \frac{1}{2}(r)\left(\frac{r\sqrt{2}}{2}\right) \\
    &= \frac{r^2\sqrt{2}}{4}
    \end{aligned}$

    There are 8 isosceles triangles, so multiply the area by 8.
    $\displaystyle 8\left(\frac{r^2\sqrt{2}}{4}\right) = 2r^2\sqrt{2}$


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  3. #3
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    A quick way to find the area the a triangle is: $\displaystyle A = \frac{{ab\sin (\phi )}}{2}$
    where $\displaystyle a~\&~b$ are the lengths of sides of the triangle and $\displaystyle \phi$ is the angle between those sides.
    That would give us of each of the triangles $\displaystyle A = \frac{{r^2 \sqrt 2 }}{4}$ at once.
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  4. #4
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    That is true, if you know trig. I had no idea if the OP learned trig or not.


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