Circular measure
Introduced| English | Meaning |
|---|---|
| arc length/ɑːk leŋθ/ | The distance along a curved part of a circle or another curve. |
| radians/ˈreɪdɪənz/ | Units of angle defined by arc length divided by radius; one full turn equals two pi of these units. |
| sector area/ˈsektə ˈeərɪə/ | The area inside two radii and their arc; it equals half r²θ when θ is in radians. |
| circular segment/ˈsɜːkjələ ˈseɡmənt/ | A region of a circle bounded by a chord and one of its arcs. |
A curved boundary needs more than a radius
- A curved window may contain an arc and several straight edges. Its perimeter is not the same as its arc length 弧长.
- First identify the angle and the region. Then combine circle formulas with triangle lengths and areas.
Define and convert radians 弧度
- One radian is the angle at the centre that cuts off an arc equal in length to the radius. Thus $\theta=s/r$; a full turn is $2\pi$ radians because its circumference is $2\pi r$.
- Use $\pi\text{ radians}=180^\circ$. Multiply degrees by $\pi/180$; multiply radians by $180/\pi$. Keep exact multiples of $\pi$ when possible, including angles larger than one turn.
$150^\circ=5\pi/6$ radians, while $7\pi/4$ radians is $315^\circ$. A turn of $450^\circ$ is $5\pi/2$ radians. One radian is about $57.3^\circ$, not $180^\circ$.
π radians equals how many degrees?
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By definition, π radians = 180°.
60° in radians is:
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60° × π/180 = 60π/180 = π/3.
Explain why a full turn is 2π radians using arc length and radius. Convert 225° and−60° to exact radians, and 11π/6 radians to degrees. For the same angle, what happens to arc length if the radius doubles?
A radian measures an angle using a ratio of lengths, not a fixed arc length in centimetres.
Use a sector forwards and backwards
- With radius $r$ and angle $\theta$ in radians, arc length is $s=r\theta$ and sector area 扇形面积 is $A=\tfrac12r^2\theta$. Length has linear units; area has square units.
- The whole sector perimeter is $2r+s$. If $s$ and $A$ are known, eliminate $\theta$: $A=\tfrac12rs$, so $r=2A/s$. Then use $\theta=s/r$.
An arc has length $8\text{ cm}$ and its sector has area $24\text{ cm}^2$. Then $r=2(24)/8=6\text{ cm}$ and $\theta=8/6=4/3$ radians. The sector perimeter is $8+2(6)=20\text{ cm}$.
These formulas require radians. For $60^\circ$, use $\theta=\pi/3$, not $60$. Set the calculator to radians before evaluating $\sin\theta$ for a radian angle.
The sector
s = rθ · A = ½r²θ
Drag the angle θ (in radians) and the radius r — the arc length and sector area follow.
A sector has radius 5 and angle 2 radians. What is the arc length (s = rθ)?
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s = rθ = 5 × 2 = 10.
A sector has radius 4 and angle 0.5 radians. What is its area (A = ½r²θ)?
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A = ½ × 4² × 0.5 = ½ × 16 × 0.5 = 4.
The formula s = rθ works when θ is measured in degrees.
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s = rθ only works when θ is in radians. Convert degrees to radians first.
A sector has arc length 12 cm and area 30 cm². Find its radius, angle in radians and full perimeter. Show how you eliminate one unknown.
An arc length is only one part of a sector perimeter.
Separate a chord, a sector and a segment
- For a minor central angle $0<\theta<\pi$, two radii and their chord form an isosceles triangle. Bisecting it gives chord length $c=2r\sin(\theta/2)$. Its area is $\tfrac12r^2\sin\theta$.
- For $0<\theta<\pi$, the minor circular segment 弓形 lies between the chord and the minor arc. Subtract triangle from sector: $A_{\text{minor}}=\tfrac12r^2(\theta-\sin\theta)$. Its perimeter is chord plus arc, not two radii plus arc.
For $r=5\text{ cm}$ and $\theta=2$ radians, the arc is $10\text{ cm}$ and sector area is $25\text{ cm}^2$. The minor segment area is $12.5(2-\sin2)\approx13.6\text{ cm}^2$. Its perimeter is $10+10\sin1\approx18.4\text{ cm}$.
A segment has r = 6 and θ = π/3 radians. The segment area is ½r²(θ − sin θ). Find it (2 dp).
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½ × 36 × (π/3 − sin(π/3)) = 18 × (1.0472 − 0.8660) = 18 × 0.1812 = 3.26.
Choose the major region explicitly
- A chord also bounds a major segment. Find its area by subtracting the minor segment from the whole circle: $A_{\text{major}}=\pi r^2-A_{\text{minor}}$.
- If the minor central angle is $\theta$, the major arc uses $2\pi-\theta$. The major segment perimeter is $r(2\pi-\theta)+c$, using the same chord as the minor segment.
Let $r=4\text{ cm}$ and chord length be $4\sqrt3\text{ cm}$. Then $\sin(\theta/2)=\sqrt3/2$. Since the minor angle is below $\pi$, $\theta=2\pi/3$.
The major arc has length $16\pi/3\text{ cm}$. Add the chord, not two radii, for the major segment perimeter.
A circle has radius 6 cm and a chord of length 6√3 cm. Find the minor central angle, the exact minor and major segment areas, and the exact major segment perimeter. Draw which region is major.
A major segment is not just a major sector: it also includes the triangle beside the chord.
Trace a compound boundary
- Suppose $OA=OB=10\text{ cm}$, $\angle AOB=1$ radian, and an inner arc has centre $O$, radius $4\text{ cm}$. Remove its sector from triangle $OAB$.
- The remaining boundary has two straight side pieces of length $6$, chord $AB=20\sin(1/2)$, and inner arc length $4$. Its area is the large triangle minus the small sector.
There is no outer circular arc on this boundary: the outer edge is the straight chord $AB$. For a different shape bounded by two concentric arcs, subtract their sector areas and include both radial gaps in its perimeter. A labelled sketch distinguishes these regions.
OA=OB=8 cm and angleAOB=1.2 radians. C and D lie on OA and OB with OC=OD=3 cm. The region is bounded by CA, chordAB, BD and the minor arcDC centred atO. Sketch it and find its perimeter and area to 3 significant figures.
Identify the region before choosing formulas; a triangle minus a sector is not a sector minus a sector.
Two concentric sectors have radii 9 cm and 5 cm and common angle 1.4 radians. The region between them is bounded by two arcs and two radial straight pieces. Sketch it and find its area and perimeter.
Subtracting areas does not mean subtracting boundary lengths.
Find an unknown radius and angle
- If a sector area is $8\theta\text{ cm}^2$, equate it to $\tfrac12r^2\theta$. Since $\theta>0$, cancellation gives $r^2=16$. Another fact about its triangle can then determine $\theta$.
- If the triangle area is $4\text{ cm}^2$, then $8\sin\theta=4$. For $0<\theta<\pi/2$, choose $\theta=\pi/6$. The minor segment area is $4\pi/3-4\text{ cm}^2$.
A different part may replace the sector-area condition. Do not reuse $r=4$ or $\theta=\pi/6$ after that condition changes. With a chord-to-radius ratio, find the angle first using $c=2r\sin(\theta/2)$, then use the triangle area to find $r$.
A sector of radius r has angle θ, where 0<θ<π/2. Its area is 18θ cm² and its triangle between the radii and chord has area 9 cm². Find r, θ and the exact minor segment area.
The interval is needed to choose between possible sine angles.
Instead of specifying a sector area, a new circle has a chord equal in length to its radius r. Its minor segment has central angle θ with 0<θ<π, and the triangle between the radii and chord has area 6 cm². Find θ, r² and the exact minor segment area. Do not reuse values from the previous task.
A changed condition requires a new equation; earlier numerical values are not permanent properties of a sector.
Subtract regions from different circles
- In the diagram, $OP=OQ=3$, $OC=4$ and $CP=CQ=5$. The required region is inside the small circle and outside the large circle. Its boundary is the small left semicircle and the large circle's short arc $PQ$.
- Let $\phi=\angle PCQ=2\tan^{-1}(3/4)\approx1.287$ radians. The large segment inside the small left semicircle has area $\tfrac12(5)^2\phi-12$: triangle $PCQ$ has base $6$ and height $4$.
Only the relevant arc of the large circle is shown. The shaded crescent is the small left semicircle minus the large circle's minor segment.
The arc and sector formulas each use their own circle's radius and central angle. Length $OC$ is not part of the outside boundary.
A shared chord can also split a two-circle region into two circular segments. Sketch which side belongs to the required region. Calculate each sector and triangle with its own radius and angle, then add or subtract those named areas.
A small circle has centreO and radius 4 units. A larger circle has centreC, radius 5 units and OC=3 units. They meet atP andQ, and PQ is a diameter of the small circle perpendicular to OC. Find the area and perimeter of the region inside the small circle but outside the large circle, to 3 significant figures. Sketch the boundary.
Use each circle’s own radius and central angle. Do not subtract two sectors when one part is a semicircle minus a segment.
Two circles each have radius 6 cm. Their centres O and C are 6 cm apart, and the circles intersect at P and Q. Sketch their overlapping region, inside both circles. Find the exact area and perimeter of this region. Show how the shared chord divides it into two segments.
An overlap region is different from a crescent outside one circle. Its boundary still contains arcs from two different centres.
Choose the region before calculating
- Convert to radians; keep length and area units distinct.
- Trace every boundary edge before adding lengths.
- Use triangle geometry to find a chord, radius or angle.
- For areas, name the sector, triangle, segment or whole circle you add or subtract.
Write an independent solution
- Draw a labelled sketch for each region problem. Show the angle, radius and formula used for every piece.
- Keep exact values until the end, then round only if requested. Check that an area is positive and smaller than its enclosing region.
A sector has radius 4 cm and angle 1.5 radians. Find its arc length and area, with units.
4 times 1.5 is 6. Half of 16 times 1.5 is 12.
A circle has radius 4 and angle 1.5 radians. What is the arc length?
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Radius times angle is 6 when the angle is in radians.