Radian measure, arcs, sectors and segments
Diperkenalkan| English | Makna |
|---|---|
| sector/ˈsektə/ | juring |
A curved window has a straight lower edge. How can we find the glass area without counting the triangle below the arc?
- A curved window has a straight lower edge. How can we find the glass area without counting the triangle below the arc?
- This lesson studies sector 扇形: The region bounded by two radii and their circular arc.
Choose the mathematical structure
- One radian is the central angle with arc length equal to the radius. A full turn is 2π radians=360°, so radians=degrees×π/180. For r>0 and a nonnegative swept angle θ in radians, arc length s=rθ and sector area A=r²θ/2. A sector perimeter is 2r+s, not s alone. Specify minor or major region.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
A sector has r=5 cm and arc 8 cm. Which quantities follow, and what boundary is missed if the perimeter is read as 8 cm?
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The inverse relations give θ=s/r=8/5=1.6 and A=rs/2=20 cm². A sector’s boundary is arc PLUS two radii: 8+5+5=18 cm. Reading the arc as the perimeter drops both straight edges.
Work through a checked case
- Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.
With r=6 cm and θ=120°=2π/3, s=4π cm and sector area=12π cm². Bisect the isosceles triangle: chord=2r sin(θ/2)=12 sin(π/3)=6√3 cm. The triangle area is r² sinθ/2=9√3 cm². The minor segment above the chord has area 12π−9√3 cm². Its boundary length is arc+chord=4π+6√3 cm. The complementary major segment has area 36π−(12π−9√3)=24π+9√3 cm². In an inverse task, radius 5 cm and arc 8 cm give θ=8/5=1.6 radians and sector area=5×8/2=20 cm².
Radian measure, arcs, sectors and segments
One radian is the central angle with arc length equal to the radius
Explain each condition before using the corresponding trigonometric formula.
Radius 5 cm and arc 8 cm: find the angle in radians.
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θ=s/r=8/5=1.6 radians.
Test a tempting shortcut
- Never put 120 directly into rθ: first convert degrees to radians. A chord is straight and generally shorter than its minor arc. Sector, segment and triangle have different boundaries. The minor-segment subtraction shown uses 0≤θ≤π; for a major segment subtract the complementary minor region from the disk.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
The perimeter of a sector consists only of its circular arc. This claim is false. Explain which definition or assumption it violates.
Find that sector area in cm².
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A=rs/2=5×8/2=20 cm².
The perimeter of a sector consists only of its circular arc.
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Never put 120 directly into rθ: first convert degrees to radians. A chord is straight and generally shorter than its minor arc. Sector, segment and triangle have different boundaries. The minor-segment subtraction shown uses 0≤θ≤π; for a major segment subtract the complementary minor region from the disk.
Interpret a new situation
- Draw the radii, chord and required boundary first. Recover θ from s/r or 2A/r² when needed. Keep π and surds exact; attach length or area units only after choosing the correct formula. In the inverse example the sector perimeter is 10+8=18 cm.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
Find that sector perimeter in cm.
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The boundary includes two radii and the arc: 5+5+8=18 cm.
Match each radian-geometry quantity to its expression.
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Degrees must be converted before rθ; the segment is found by subtracting the triangle from the sector - the two regions have different boundaries.
Use this in your course
- 7357 · A-level · E. Match the target tier and specification before assigning extensions.
- Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.
- The checked inverse example stays radius 5 cm and arc 8 cm. The later problems use other lengths and angles. Decide sector or segment before you substitute. A major segment is the disk minus the minor segment on the same chord.
The region bounded by two radii and their circular arc. Choose the relationship, show the method, check its assumptions and interpret the result.
Radius 9 cm and angle 2/3 radian. Find the arc length in cm.
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s=rθ=9×(2/3)=6 cm. These are not the radius-5 cm, arc-8 cm values.
Radius 8 cm and angle 1.5 radians. Find the sector area in cm².
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A=½r²θ=½×64×1.5=48 cm².
A sector has area 50 cm² and radius 10 cm. Find the angle in radians.
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½r²θ=50 gives ½×100×θ=50, so θ=1 radian. The radius and the area are not the worked 5 cm and 8 cm pair.
An arc is 12 cm and the angle is 0.4 radians. Find the radius in cm.
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r=s/θ=12/0.4=30 cm.
A garden bed is bounded by two straight edges from a sprinkler and the watered arc. No chord is part of the boundary. Which area is asked?
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Two radii and the arc bound a sector. The segment formula subtracts the triangular part up to a chord, and that chord is not the boundary here.
For 0≤θ≤π the minor segment is ½r²(θ−sinθ). Which quantity is the major segment on the other side of the same chord?
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The major segment and the minor segment together make the disk. The major sector leaves out the triangle that sits in the minor sector, so ½r²(2π−θ) is not the major segment. The minor sector is smaller than the major segment.
A wiper sweeps a sector of radius 40 cm through 0.5 radians. Inside that sector a triangular patch of 200 cm² stays dry. The wet area is the sector minus that triangle. Find the wet area in cm².
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Sector area=½×40²×0.5=400 cm². Wet area=400−200=200 cm². The triangle is given, so this is not the radius-5 cm, arc-8 cm example.