Visual walkthrough — Circles — centre, radius, diameter, chord, arc, sector, segment

This page rebuilds one result from the parent note on Circles from the ground up, in pictures. The result:

$A_{\text{segment}}=\frac{1}{2}{r}^2(\theta -\sin \theta )$

We will not assume you know what a sector, a chord, a radian, or a sine is. Every one of those is a picture before it is a symbol. By the end you will see why this formula must be true — and why the little $-\sin \theta$ at the end is not a mystery but a triangle we chose to remove.

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Step 1 — Draw the circle and name its one number, $r$

WHAT. We draw a circle. We mark its centre — the anchor point everything is measured from — and call it $O$. We pick any point on the rim and draw a straight spoke from $O$ to it. That spoke's length is the radius, written $r$.

WHY. Before we can talk about slices or slivers, we need the one measurement a circle is built from. A circle is defined as "all points the same distance $r$ from $O$". So $r$ is the only length that exists here until we invent more. Every area we compute will be some number times $r^2$ — because area is a length times a length, and $r$ is our only length.

PICTURE.

If you want the full family of parts (diameter, chord, arc) revisited, that is the parent's job; here we need only $r$.

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Step 2 — Open two spokes by an angle $\theta$ (in radians)

WHAT. We draw a second spoke from $O$, so now two radii fan out. The amount they are opened apart is the central angle, written $\theta$. We will measure $\theta$ in radians, not degrees.

WHY radians and not degrees? A radian is defined so that the length of arc your spoke sweeps equals $r\theta$ — clean, no conversion factor. Degrees would force us to drag around a factor of $\frac{\pi }{180}$ in every line. Choosing radians is choosing the unit in which the formulas are simplest. (More on this in Radian Measure.)

PICTURE.

$\underset{\text{angle between the two spokes, in radians}}{\underbrace{\theta }}$

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Step 3 — The two spokes plus the arc cut out a sector

WHAT. The region trapped between the two radii and the curved rim between their tips is a sector — the pizza slice.

WHY start here? Because the segment is hiding inside the sector. If we can measure the whole slice first, we only have to subtract the pointy part later. We build from the bigger, simpler shape down to the sliver.

PICTURE.

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Step 4 — Measure the sector by fair sharing of the whole circle

WHAT. We find the sector's area by asking: what fraction of the whole circle is this slice?

WHY this trick? The whole circle's area is a known fact, $\pi r^2$ (built by stacking thin rings — see the area-of-a-circle story, or Integration for the ring-stacking done formally). A sector is just a fraction of that whole. A full circle is $2\pi$ radians of angle; our slice is $\theta$ radians. So the slice is the fraction $\frac{\theta }{2\pi }$ of the pie.

PICTURE.

$A_{\text{sector}}=\underset{\text{what fraction of the full turn}}{\underbrace{\frac{\theta }{2\pi }}}\times \underset{\text{area of the whole circle}}{\underbrace{\pi r^2}}=\frac{1}{2}{r}^2\theta$

Term by term: $\frac{\theta }{2\pi }$ is "how big a slice"; $\pi r^2$ is "the whole pie"; the $\pi$ cancels, leaving the tidy $\frac{1}{2}{r}^2\theta$.

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Step 5 — Split the sector: pointy triangle + flat segment

WHAT. We drop the straight chord connecting the two spoke-tips. That chord slices the sector into exactly two pieces:

• a triangle $OAB$ (the pointy part, corner at the centre),
• a segment (the flat sliver against the rim).

WHY. This is the key move. The segment is awkward — one straight edge, one curved edge. But the triangle is easy, and:

$A_{\text{segment}}=A_{\text{sector}}-A_{\text{triangle}}$

We already own the sector (Step 4). If we can measure the triangle, we are done by subtraction.

PICTURE.

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Step 6 — Measure the triangle: where $\sin \theta$ is born

WHAT. The triangle $OAB$ has two sides equal to $r$ (both are spokes), meeting at the angle $\theta$. We need its area.

WHY $\sin$ and not something else? Area of a triangle is $\frac{1}{2}(\text{base})\times (\text{height})$. Take one radius $OA$ as the base — length $r$. Now we need the height: how far the far tip $B$ stands above that base line. Drop a straight-down line from $B$ to the base. Look at the right triangle it makes: its slanted side is the other radius (length $r$), and the angle at $O$ is $\theta$.

The sine of an angle is defined as "the far side (opposite the angle) divided by the slanted side (hypotenuse)". So:

$\sin \theta =\frac{\text{height}}{r}⟹\text{height}=r\sin \theta$

That is exactly why $\sin$ shows up: it is the tool that converts "an angle $\theta$" into "how tall the triangle stands". No other function answers that question.

PICTURE.

$A_{\text{triangle}}=\frac{1}{2}\times \underset{\text{base}}{\underbrace{r}}\times \underset{\text{height}}{\underbrace{r\sin \theta }}=\frac{1}{2}{r}^2\sin \theta$

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Step 7 — Subtract, and read the final formula

WHAT. Put Steps 4 and 6 into the subtraction from Step 5.

WHY. This is the payoff — the segment is (slice) minus (triangle):

$A_{\text{segment}}=\underset{\text{sector, Step 4}}{\underbrace{\frac{1}{2}{r}^2\theta }}-\underset{\text{triangle, Step 6}}{\underbrace{\frac{1}{2}{r}^2\sin \theta }}=\frac{1}{2}{r}^2(\theta -\sin \theta )$

The common $\frac{1}{2}{r}^2$ factors out. What is left inside the bracket is a race between two numbers: $\theta$ (the angle itself) and $\sin \theta$ (the height-ratio). The gap $\theta -\sin \theta$ is the sliver.

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Step 8 — Every case, including the weird ones

A formula you can't trust at the edges is a formula you don't understand. Check the limits.

PICTURE.

• Tiny angle, $\theta \to 0$. The two spokes nearly overlap; there is barely any sliver. For small angles $\sin \theta \approx \theta$, so $\theta -\sin \theta \approx 0$. Area $\to 0$. ✓ Matches the picture — no slice, no sliver.
• Half circle, $\theta =\pi$. The chord becomes a diameter; the segment is a semicircle. Here $\sin \pi =0$, so $A=\frac{1}{2}{r}^2(\pi -0)=\frac{1}{2}\pi r^2$ — exactly half the circle's area $\pi r^2$. ✓
• Right angle, $\theta =\frac{\pi }{2}$. $\sin \frac{\pi }{2}=1$, so $A=\frac{1}{2}{r}^2\left(\frac{\pi }{2}-1\right)$. With $r=8$ this is $\frac{1}{2}(64)(\frac{\pi }{2}-1)=16\pi -32\approx 18.27$ — the parent's Example 3. ✓
• Reflex / major segment, $\theta >\pi$. Now the sliver is the big piece. The triangle's "height" $r\sin \theta$ goes negative (the tip drops below the base), which is the formula's honest way of saying the triangle now sits on the other side and must be added, not subtracted. The single formula $\frac{1}{2}{r}^2(\theta -\sin \theta )$ handles it automatically — e.g. $\theta =\frac{3\pi }{2}$ gives $\frac{1}{2}{r}^2(\frac{3\pi }{2}+1)$, a large area, as it should. ✓
• Full circle, $\theta =2\pi$. $\sin 2\pi =0$, so $A=\frac{1}{2}{r}^2\cdot 2\pi =\pi r^2$ — the whole circle. ✓ The formula closes the loop.

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The one-picture summary

The whole derivation on one canvas: whole circle $\pi r^2$ → take the fraction $\frac{\theta }{2\pi }$ to get the sector $\frac{1}{2}{r}^2\theta$ → drop the chord to split off the triangle $\frac{1}{2}{r}^2\sin \theta$ → subtract to keep the segment $\frac{1}{2}{r}^2(\theta -\sin \theta )$.

Recall Feynman retelling — say it to a friend with no maths

Imagine a pizza. First I ask: how big is one slice? Well, the whole pizza has some size, and my slice is just a fraction of the turn — if I opened my slice a quarter of the way around, my slice is a quarter of the pizza. That fraction times the whole gives me the slice's area, and it comes out to half of "radius times radius times the angle". Now, a segment is that slice with the pointy tip broken off along a straight cut. So I measure the pointy triangle: it's got two sides that are both the pizza's radius, meeting at my angle. Its area is half of "radius times radius times the sine of the angle" — where "sine" is just the trick that tells me how tall the triangle stands for a given opening angle. Finally I take the slice and remove the triangle, and the leftover flat sliver is half of "radius times radius times (the angle minus its sine)". When the slice is thin, angle and sine almost match, so the sliver is nearly nothing — which is exactly what my eyes see. When I open it to a straight line, the sliver becomes a perfect half-pizza. The one formula quietly gets every case right.

Recall Rapid self-check

Sector area for angle $\theta$ radians, radius $r$? ::: $\frac{1}{2}{r}^2\theta$ Triangle (two radii, included angle $\theta$) area? ::: $\frac{1}{2}{r}^2\sin \theta$ Segment area? ::: $\frac{1}{2}{r}^2(\theta -\sin \theta )$ Why does $\sin$ appear at all? ::: It turns the angle $\theta$ into the triangle's height $r\sin \theta$. What must $\theta$ be measured in? ::: Radians. Segment area when $\theta =\pi$? ::: A semicircle, $\frac{1}{2}\pi r^2$.

Related deeper threads: Angle Properties in Circles, Circle Theorems, Circle Equations, Pythagorean Theorem (used to find the chord's length from the triangle).