# 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}

## Proof 3: Circle circumference and arc length

The circumference formula is quick to figure out since by definition: $\text{circumference}=\pi\;\text{diameter}$ (as explained in the Natural constants and pi skill) and since the diameter…

## Proof 11: Sphere and ellipsoid volume

A sphere can be cut into slices. Each slice is almost a disk, or a very flat cylinder, except for the curved edge. If we…

## Proof 9: Rhombus area

A rhombus actually consists of four right-angled triangles, because the diagonals $d_1$ and $d_2$ are perpendicular. Those triangles are pair-wise equal. The rhombus area is…

## Proof 10: Parallelogram and trapezium area

A trapezium consists of a rectangle and two weird triangles at the ends. \begin{align}A_\text{trapezium}&= A_\text{rectangle}+ A_{\text{triangle } a}+ A_{\text{triangle } b}\\~&=\underbrace{lw_1}_\text{rectangle}+\underbrace{\frac12 w_al}_{\text{rectangle } a}+\underbrace{\frac12 w_b…

## Proof 30: Sphere’s moment-of-inertia

Solid sphere (ball) spinning about an axis through its centre-of-mass Let’s cut the ball into many thin slices. Each slice has a small thickness \$\mathrm…