Proof 4: Circle and circle sector area

A pizza slice’s edge is the pizza’s radius $r$.

The slice is almost a triangle if it wasn’t for the rounding. That rounding has a length $w$. Cutting a pizza in many slices means a smaller and smaller rounding. The smaller and smaller $w$ looks more and more like a straight line and the slice more and more like a triangle. If the $n$ slices it is cut into is a very, very large number, then we can work with each slice as was it a triangle.

The whole pizza area is made up of all the slices’ areas:

\begin{align} A_\text{circle}&=\underbrace{A_1+A_2+A_3+\cdots+a_n}_{n\text{ terms}}\\ ~&=\underbrace{\frac12rw+ \frac12rw + \frac12rw +\cdots+ \frac12rw }_{n\text{ terms}}\qquad \leftarrow \text{ Triangle area is } A=\frac12 rw\\ ~&=n\frac12rw\\ ~&=\frac12r\,\text{circumference} \qquad \leftarrow \text{ All } w \text{-sides make up the full round, circumference} =nw\\ ~&=\frac1{\cancel 2}r\,\cancel{2}\pi r \qquad \leftarrow \text{ circumference} =2\pi r\\ ~&=\pi r^2\quad_\blacksquare \end{align}

When cut into regular pieces of angle $\theta$, each piece covers a portion of the total pizza area. We call such piece a sector of the circle. If the angle is, say, one sixth of the full round, then its area must also cover one sixth of the total area. Meaning, the ratios are the same:

\begin{align} \frac{A_\text{sector}}{ A_\text{circle} }=&\frac \theta { \theta_\text{circumference} }\quad\Leftrightarrow\\ A_\text{sector}=&\frac \theta { \theta_\text{circumference} } A_\text{circle}\\ ~ =&\frac \theta { 2\cancel\pi } \cancel \pi r^2 \qquad \leftarrow \text{ circumference} =2\pi r \\ ~=&\frac12 \theta r^2 \quad_\blacksquare \end{align}