Arcs, sectors and segments: what are we measuring?

A circular plot needs two different purchases: edging along its boundary and covering for its interior. Both numbers may contain π, but one is in metres and the other in square metres. How do we keep them separate?

A circle is the boundary; a disk includes the interior and boundary. An arc has length; sectors and segments have area. Their formulas depend on the fraction of a complete turn. We use radii and chords, central angles and area dissection.

Why pi appears twice

Scaling an ideal circle by k scales both diameter d and circumference C by k. The ratio C/d stays fixed. We call this common ratio π:

$\pi=\frac{C}{d},\qquad C=2\pi r,\qquad d=2r.$

π≈3.14159. Both 3.14 and 22/7 are approximations. String around a tin helps estimate the ratio, but tension, string thickness and the object’s shape introduce error. Such an experiment does not prove pi’s irrationality. AMSI, circumference.

Disk area S=πr² is also connected to boundary length. Circumscribe a regular n-gon about the disk. Each central triangle has altitude r, so its total area is Sₙ=Pₙr/2. As the number of sides grows without bound, Pₙ approaches C and Sₙ approaches disk area. Thus S=Cr/2=πr². This uses a limit, rather than claiming a finite polygon already is a disk.

One arc, three objects

Two radii cut off a sector, bounded by those radii and the arc. A chord with the same arc bounds a segment. For a central angle smaller than 180°, the sector contains the triangle between radii and chord; the minor segment excludes it.

Arc: length

Sector: area

Segment: area

The panels share scale and the same arc AB: r=3m, α=120°=2π/3radians. The orange line in panel1 has length2πm. The shaded sector in panel2 has area3πm²; its boundary includes OA and OB. The shaded segment in panel3 is bounded by arc and chord AB; triangle OAB below the chord has been subtracted. Dashed radii are construction lines, not segment boundaries. The artistic preview is separate from this calculated diagram.

Let central angle α be in degrees, 0≤α≤360°. The fraction of a turn is f=α/360°. Arc length is ℓ=f·2πr and sector area is S=f·πr². They share a fraction, but one scales with r and the other with r².

For r=3 m and α=120°, we take one-third of a turn. Arc length ℓ=2π≈6.283 m; sector area S=3π≈9.425 m². To fence the entire sector, include two radii: P=2π+6≈12.283 m. The arc alone omits its straight edges.

Radians remove a factor

Radian measure θ is arc length divided by radius: θ=ℓ/r. One radian corresponds to an arc as long as r. A full circumference is 2πr, so a full turn is 2π radians and 180° is π radians.

$\theta=\frac{\alpha}{180\text{°}}\pi,\qquad \ell=r\theta,\qquad S_{\text{sector}}=\frac12r^2\theta.$

The last two formulas require θ in radians. Our 120° is 2π/3 radians. Substituting 120 instead of 2π/3 would give a very different arc. Check DEG/RAD when calculating sines as well. OpenStax, radians and sectors.

A segment: subtract the triangle

For a minor segment with 0<θ≤π, subtract from the sector the triangle with sides r,r and included angle θ. Its area is r²sinθ/2 by G018. Thus:

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

Here θ and the sine argument both use radians. For r=3,θ=2π/3, triangle area is 9√3/4≈3.897 and segment area is 3π−9√3/4≈5.528 m². It is smaller than the sector, as expected. Chord length is 2r sin(θ/2)=3√3. The segment’s complete boundary is arc plus chord, rather than two radii. AMSI, sectors and segments.

The rest of the diskOptional · safe to skip

Finding the major segment

The same chord also cuts off a major segment. Its area is disk area minus minor-segment area: 9π−(3π−9√3/4)=6π+9√3/4≈22.747 m². It includes the large 240° sector plus the small triangle, so we add triangle area here.

The single formula r²(θ−sinθ)/2 also works for π<θ<2π when θ describes the major arc: sinθ is negative, so subtraction becomes addition. Do not replace the signed sine by a positive triangle area.

Scale and a materials estimate

At the same angle, doubling radius doubles arc, chord and perimeter; sector and segment areas quadruple. Metres of edging cannot be added to square metres of covering. Identify the required boundary or region, calculate using π, then round at the end. Material allowance is a separate decision, rather than a hidden term in a geometric formula.