Practice Maths

L24 — Applications of Circle Geometry

Key Terms

Sector
A “pie slice” region bounded by two radii and an arc. The central angle θ determines its size.
Arc length
The distance along the arc: l = (θ/360) × 2πr. It is a fraction of the full circumference.
Sector area
The area of the sector: A = (θ/360) × πr². It is the same fraction of the full circle area.
Segment
The region between a chord and the arc it cuts off. Segment area = sector area − triangle area.
Circumscribed circle
A circle passing through all vertices of a polygon. The centre is the circumcentre.
Inscribed circle (incircle)
A circle touching all sides of a polygon from inside. The centre is the incentre. For a right triangle: radius = (a + b − c) / 2.

Arc Length and Sector Area

For a sector with radius r and central angle θ (in degrees):

QuantityFormula
Arc length l = (θ/360) × 2πr
Sector area A = (θ/360) × πr²
Segment area Sector area − Triangle area
r r θ arc = (θ/360)×2πr O segment

Circumscribed and Inscribed Circles

Circumscribed circle: passes through all vertices of a polygon. The centre is the circumcentre.

Inscribed circle (incircle): touches all sides of a polygon. The centre is the incentre.

For a right triangle with legs a, b and hypotenuse c: incircle radius = (a + b − c) / 2.

Hot Tip: Arc length and sector area use the same fraction θ/360 of the full circle. A shortcut: sector area = ½ × arc length × r. This is always quicker than substituting into both formulas separately.

Arc Length

An arc is a fraction of the full circumference. The fraction is θ/360.

Worked Example 1 — Arc length

Sector: r=8 cm, central angle 135°. Find the arc length.

l = (135/360) × 2π(8) = (3/8) × 16π = 6π ≈ 18.85 cm.

Sector Area

Worked Example 2 — Sector area

Sector: r=10 cm, central angle 72°. Find the sector area.

A = (72/360) × π(100) = (1/5) × 100π = 20π ≈ 62.83 cm².

Segment Area

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

Worked Example 3 — Segment area

Sector: r=6 cm, central angle 90°. Find the minor segment area.

Sector area = (90/360)×π(36) = 9π. Triangle area = ½(6)(6) = 18 cm².

Segment = 9π − 18 ≈ 28.27 − 18 = 10.27 cm².

Real-World Applications

Circle geometry appears in architecture (arches, domes), engineering (gears, wheels), navigation, and design. Always identify which theorem or formula applies before computing.

  1. Arc length. Fluency

    Give answers in terms of π and as a decimal (2 d.p.).

    • (a) r = 6 cm, θ = 60°.
    • (b) r = 10 m, θ = 90°.
    • (c) r = 5 cm, θ = 216°.
    • (d) Arc length = 12π cm, r = 9 cm. Find θ.

  2. Sector area. Fluency

    Give answers in terms of π and as a decimal (2 d.p.).

    • (a) r = 8 cm, θ = 45°.
    • (b) r = 12 m, θ = 120°.
    • (c) r = 5 cm, θ = 300°.
    • (d) Sector area = 30π cm², θ = 120°. Find r.

  3. Segment area. Fluency

    Give answers to 2 decimal places.

    • (a) r = 10 cm, θ = 90°. Find the minor segment area.
    • (b) r = 6 cm, θ = 60°. Find the minor segment area. (Triangle is equilateral.)
    • (c) r = 8 cm, θ = 120°. Find the minor segment area.
    • (d) r = 5 cm, θ = 144°. Find the major segment area.

  4. Perimeter of sectors and segments. Fluency

    • (a) Sector r=7 cm, θ=80°. Find the perimeter (arc + two radii).
    • (b) Sector r=12 m, θ=150°. Find the perimeter.
    • (c) Segment r=9 cm, θ=120°. Find the perimeter (arc + chord).
    • (d) A semicircle has diameter 20 cm. Find its perimeter and area.

  5. Sector and segment from a diagram. Understanding

    The diagram shows a circle with centre O and radius 12 cm. The shaded region is a minor segment cut off by chord AB. The central angle AOB = 60°.

    60° 12 cm O A B shaded segment
    • (a) Find the arc length AB (the minor arc).
    • (b) Find the length of chord AB. (Hint: the triangle OAB is equilateral when θ=60°.)
    • (c) Find the area of the minor sector OAB.
    • (d) Find the area of the shaded minor segment.

  6. Inscribed and circumscribed circles. Understanding

    • (a) An equilateral triangle has side length 6 cm. Find the circumradius (radius of circumscribed circle). Use R = a/√3.
    • (b) A right triangle has legs 6 cm and 8 cm. Find the circumradius.
    • (c) A right triangle has legs 5 cm and 12 cm. Find the inradius using r = (a+b−c)/2.
    • (d) A square with side 10 cm is inscribed in a circle. Find the radius of the circle.

  7. Mixed circle applications. Understanding

    • (a) A windscreen wiper sweeps a sector of radius 45 cm and angle 110°. Find the area swept.
    • (b) A pizza (radius 15 cm) is cut into 8 equal slices. Find the arc length of one slice’s curved edge.
    • (c) A running track has a semicircular end with inner radius 36 m and outer radius 37.22 m. Find the area of the semicircular lane.
    • (d) A clock hand of length 8 cm sweeps from 12 to 3 (90°). Find the area swept and the distance the tip of the hand travels.

  8. Circles and triangles. Understanding

    • (a) A circle of radius 5 cm has a chord that subtends a central angle of 120°. Find the chord length.
    • (b) A chord of length 8 cm subtends an angle of 90° at the centre. Find the radius.
    • (c) An isosceles triangle is inscribed in a circle of radius 10 cm. The apex angle is 40°. Find the base of the triangle.
    • (d) A chord of length 10 cm subtends an inscribed angle of 30° at a point on the major arc. Find the radius of the circle.

  9. Arch design. Problem Solving

    An architect designs a doorway arch as a circular arc. The opening is 2 m wide and the arc rises 0.5 m above the chord (the “rise”).

    • (a) Let r be the radius of the circle. Using the perpendicular from the centre to the chord, show that r = (1 + 0.25×r²−1)/0.5… More precisely: half-chord = 1 m, sagitta (rise) = 0.5 m. Use the sagitta formula: sagitta = r − √(r²−a²), where a = 1 m, to find r.
    • (b) Find the central angle subtended by the arch.
    • (c) Find the arc length of the arch.
    • (d) Find the area of the segment (the filled region under the arch).

  10. Gear wheels. Problem Solving

    Two interlocking gear wheels: Gear A has radius 8 cm and Gear B has radius 6 cm. Their centres are 14 cm apart.

    • (a) Verify the gears mesh correctly (the circles are externally tangent). What theorem guarantees the tangent line at their contact point is perpendicular to the line of centres?
    • (b) Gear A rotates 45°. Find the arc length on Gear A that moves past the contact point.
    • (c) By how many degrees does Gear B rotate when Gear A rotates 45°? (Arc lengths must be equal at the contact point.)
    • (d) If Gear A makes 1 full rotation per second, how many rotations per second does Gear B make? What is this ratio called?
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