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Math Help - 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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    OC^2 + AC^2 = OA^2

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

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

    There are 8 isosceles triangles, so multiply the area by 8.
    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: A = \frac{{ab\sin (\phi )}}{2}
    where a~\&~b are the lengths of sides of the triangle and \phi is the angle between those sides.
    That would give us of each of the triangles 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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