Thread: Help request on the pythagorean Theorem 2.

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

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.
$\displaystyle 8\left(\frac{r^2\sqrt{2}}{4}\right) = 2r^2\sqrt{2}$


01

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.

That is true, if you know trig. I had no idea if the OP learned trig or not.


01