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Area of a segment

Find minor and major segment areas using sectors and triangles. Recover angles from chords or sagitta, form area-ratio equations and solve curved cross-section problems.

Before you startSector area, triangle area, inverse sine and Pythagoras

01 / A chord changes the boundary

A segment is the region between a chord and an arc.

A sector reaches the centre along two radii. A segment stops at the straight chord between the arc’s endpoints. For a minor central angle, remove the triangle from the sector to leave the minor segment.

Minor segment area = sector area − triangle area

Use the model to shade each piece. At θ = π the chord is a diameter: the triangle has zero area and the two segments are equal semicircles.

Sector, triangle, then segmentExplore

Selected area ≈4.566371 square units.

Sector ≈12.566371; triangle =8; minor segment ≈4.566371; major segment ≈45.699112.

Choose a region to shade it. The two segment areas add to the full circle. The formula uses the minor central angle; choose the major region by taking the complement.

Radius 4Minor angle in steps of π/6 π/2Region

Watch the triangle leave the sector

Pause, replay or seek freely. The notes explain the same idea and stay in view.

02 / Derive the segment formula

Both the sector and triangle use the same radius and angle.

Sector area = ½r²θ
Triangle area = ½r²sin θ
Minor segment area = ½r²(θ − sin θ)

Use the minor central angle 0 ≤ θ ≤ π in radians.

The triangle formula is ½ab sin C with a = b = r and included angle C = θ. Its sine term is dimensionless, just as θ is. Both parts must use the same angle unit; do not calculate θ in degrees and then put that number into the radian formula.

For 0 < θ < π, sin θ < θ, so the area is positive. A negative minor-segment area is a signal to check angle mode or subtraction order.

03 / Keep the exact sector and triangle areas

A fractional multiple of π often gives an exact surd result.

Radius 6 cm and minor angle π/3Worked example

Sector =½ ×36 ×π/3 =6π

Area is measured in cm².

Triangle =½ ×36 ×sin(π/3) =9√3

The triangle is not the same fraction of the circle as the sector.

Minor segment =6π −9√3 cm²

Round only if a decimal is requested.

Radius 4 cm and angle π/2Worked example

Sector 4π; triangle 8

sin(π/2) =1.

Minor segment =4π −8 cm²

This is approximately 4.56637 cm², matching the default model.

04 / Take the complement for a major segment

The straight chord divides the whole circle into two regions.

Major segment area = πr² − minor segment area

Radius 6 cm and minor angle π/3Worked example

Full circle area =36π

The minor segment is 6π −9√3.

Major segment =30π +9√3 cm²

Subtract the entire minor-segment expression.

You can also use the reflex angle φ =2π −θ in ½r²(φ −sin φ). Since sin φ =−sin θ, the triangle contribution is added. Subtracting a positive triangle area from the reflex sector would give the wrong result. The full-circle complement is usually easier to interpret.

05 / Find the angle and centre-to-chord distance

Bisect the chord to create a right triangle.

θ =2arcsin[c/(2r)]
d =√[r² −(c/2)²] =r cos(θ/2)

c is the chord length; d is its perpendicular distance from the centre. Use the minor angle.

Radius 5 cm and chord 8 cmWorked example

Half-chord 4 cm gives d =√(25 −16) =3 cm

The perpendicular from the centre bisects the chord.

θ =2arcsin(4/5)

Keep the unrounded angle when finding the area.

Triangle area =½ ×8 ×3 =12 cm²

This avoids evaluating sin θ separately.

Minor segment =25arcsin(4/5) −12 cm²

The inverse sine returns radians. Numerically this is about 11.18238 cm².

06 / Recover a radius from segment height

The sagitta is the distance from the chord to the arc at its midpoint.

For a minor segment, let its chord length be c and its height be h>0. The centre-to-chord distance is r−h. A right triangle connects this distance, the half-chord and the radius.

Derive a useful radius formulaWorked example

(c/2)² +(r −h)² =r²

Apply Pythagoras.

c²/4 −2rh +h² =0

Expand and cancel r².

r =(c² +4h²)/(8h)

Divide by 2h after rearranging.

For c =8 cm and h =2 cm, r =(64 +16)/16 =5 cm. This recovers the previous example. A semicircle has h =r; a minor segment has h<r. State which side of the chord the given height describes before using the geometry.

07 / Turn an area ratio into an equation

The radius can cancel, leaving only the angle.

A minor segment occupies one fifth of its circleWorked example

½r²(θ −sin θ) =(1/5)πr²

r>0, so divide by r².

θ −sin θ =2π/5

This is an equation for the minor central angle.

The associated major segment occupies four fifths

Its area is four times the minor segment’s area.

The exact result may be a trigonometric equation rather than a familiar angle. Do not replace sin θ by θ unless a justified approximation is explicitly requested: that would make the left-hand side zero and destroy the area information.

08 / A crescent between two arcs

Use a common chord to separate the pieces.

A semicircle has diameter 8 cm. A second arc with the same endpoints belongs to a circle of radius 5 cm whose centre is 3 cm above the chord. The second arc lies inside the semicircle. Find the crescent between the two arcs.

Subtract the inner segmentWorked example

Semicircle area =½π ×4² =8π

Its radius is 4, half the common chord.

Inner segment =25arcsin(4/5) −12

This is the segment derived above.

Crescent =8π −25arcsin(4/5) +12 cm²

The final area is positive and smaller than8π.

The outer arc is a semicircle of radius 4. The inner arc belongs to the radius 5 circle. Subtract its small segment from the semicircle.

09 / A curved cut-out from a rectangle

Subtract the segment, not the whole sector.

A rectangular cross-section is 6 cm wide and 8 cm high. A minor circular segment is cut into its left edge: the chord is that full 8 cm edge and the circle radius is 5 cm. The cut-out extends 2 cm into the rectangle, leaving positive width.

Find the remaining cross-sectional areaWorked example

Rectangle area =6 ×8 =48 cm²

The circle centre is 3 cm outside the left edge.

Removed segment =25arcsin(4/5) −12 cm²

The same chord/radius pair gives the same small segment.

For radius 4 and minor angle π/2, a student finds the major-segment area as12π−8. Correct it.

Hint

The reflex sector must include the small triangle on the other side of the centre-to-endpoint rays.

Worked solution

Full circle 16π minus minor segment(4π−8) gives12π+8. The triangle term is added, not subtracted.

11 / Recap

A curved region becomes manageable when its pieces are named.

Minor segment = sector − triangle.
Use radians in½r²(θ−sinθ).
Major segment = full circle − minor segment.
A half-chord triangle gives angle, distance and sagitta.
Cancel common squared radii when forming area-ratio equations.
Subtract only the part actually removed from a composite shape.

Section 1 of 11 · A chord changes the boundary