Is this a valid proof for the area of a circle? My teacher challenged my class to prove that the area is
$$A=\pi r^2.$$
We recently learned about Riemann sums, so I thought it would be possible to apply them to them to deriving the formula for the area of the circle.  I know similar proofs exist, but this is one that I genuinely came up with on my own and I am wondering if it is valid. Please tell me if there is anything invalid about this proof or how it can be improved.
Imagine splitting a circle into an infinite number of isosceles triangles, where two legs extend from a vertex at the center of the circle to the edge of the circle.  The central angle that each triangle makes can be represented as $\frac{2\pi}{n}$, where $n$ is the number of triangles in the circle.
The area of a triangle is $A=\frac{1}{2}ab\sin{C}$.  Since the legs of each of the triangles extends from the center of the circle to the edge, that means that $a=b=r$, the radius of the circle.  Therefore, the summation of every triangle in the circle as the number of triangles approaches infinity can be represented as:
$\lim_{ n\to\infty} \sum_{i=1}^{n} \frac{1}{2}r^2\sin(\frac{2\pi}{n})$
which can be rewritten as:
$\lim_{ n\to\infty} \frac{1}{2}r^2n\sin(\frac{2\pi}{n}) = 2\pi(\frac{1}{2})r^2 = \pi r^2$
 A: Nice proof! The only property of $\pi$ and the circle you seem to be relying on is that a central angle of $\alpha$ radians corresponds to an arclength of $\alpha 2\pi r$, that is, you rely on the formula for the circumference of a circle, but not on the formula for its area, so I don't see a circularity here. Note though that you've given an argument which does not use the standard mathematical definition of area given by definite integrals, since you've used specific triangles instead of arbitrary rectangles, but I wouldn't count that against your proof. I like your trick as it makes things simple computationally.
You can also use the standard approach via Riemann sums, where we fit rectangles inside. Let $f(x)=\sqrt{r^2-x^2}$. The graph of this function is the upper part of a circle of radius $r$ centered at the origin. When we fit an infinite number of rectangles we get $\int_{-r}^r \sqrt{r^2-x^2}\,dx$ which can be calculated using the fundamental theorem of calculus and the substitution $x=r\sin t$:
\begin{align*}
\int_{-r}^r \sqrt{r^2-x^2}\,dx &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{r^2-r^2\sin^2t}\cdot r\cos t \, dt = r^2 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^2 t\,dt\\
&= \frac{r^2}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}1+\cos(2t)\,dt=\frac{r^2}{2}\left[t+\frac{\sin(2t)}{2}\right]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\\
&= \frac{\pi r^2}{2}
\end{align*}
so that the area of the whole circle is $2\cdot\frac{\pi r^2}{2}=\pi r^2$.
Your idea of taking isosceles triangles, instead of the more standard rectangles used in integration, utilizes the symmetry of the circle and definitely makes the calculations easier in this case!
