# Thread: Help request on the pythagorean Theorem 2.

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

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

3. 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.

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

01