Help request on the pythagorean Theorem 2.

**yeongil** · Jun 18th 2009, 02:32 PM

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

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

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

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

01

**Plato** · Jun 18th 2009, 03:13 PM

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.

**yeongil** · Jun 18th 2009, 04:01 PM

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

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