Circles & Sectors

SPEC G17 G18 — GEOMETRY & MEASURES

Circumference, area, arc length, sector area, composite shapes and the discipline of keeping answers in exact form using π.

What You'll Learn

Circle questions appear on almost every paper — usually as 2–4 mark setters, sometimes inside a longer composite or annulus problem. The formulae are short and the technique is mechanical, so the marks here are easy to bank if you stay disciplined with π.

By the end of this lesson, you will be able to:

Recall and apply the four circle formulae: C = 2πr = πd, A = πr², arc length = (θ/360) × 2πr, sector area = (θ/360) × πr²

Halve a diameter to a radius before using any formula that needs r

Calculate arc lengths and sector areas for any sector angle, including semicircles and quarter circles

Find areas and perimeters of composite shapes that combine rectangles with semicircles or quarter circles

Find the area of an annulus (ring): π(R² − r²)

Reverse-engineer a radius (or sector angle) from a given arc length, sector area or circumference

Keep answers in exact form using π — never replacing it with 3.14

Why It's High Priority

"Give your answer in terms of π" appears on virtually every paper. The marks lost on circle questions are nearly always the same: using the diameter where the formula wants a radius, forgetting (1/360)×θ on a sector, or replacing π with 3.14 too early and losing the exact-form mark.

Core Circle Formulae

You must memorise both — these are not on the formula sheet at GCSE.

Circumference = 2πr = πd

Quick Identification

r = distance from centre to edge. d = full distance across through the centre. d = 2r, so r = d/2.

Radius or Diameter — Halve First

Both circle formulae take the radius. If a question gives you a diameter, halve it before substituting.

The "Use d as r" Trap

Given diameter 10 cm, the area is not π × 10² = 100π. The radius is 10 ÷ 2 = 5, so the area is π × 5² = 25π cm². Using d in place of r quadruples the area.

Two Versions of Circumference

C = 2πr and C = πd are the same formula (because d = 2r). Pick the one that matches what the question gave you — it saves the halving step:

If r is given: C = 2πr is fastest

If d is given: C = πd is fastest

For Area, you must always go via the radius first

01Worked Example — Circle (radius given)

A circle has radius 7 cm. Find (a) its circumference and (b) its area. Give exact answers in terms of π. (3 marks)

MMethod. Use C = 2πr and A = πr² with r = 7. Keep π as a symbol throughout — the question asks for exact form.

CCalculate.

(a) C = 2π × 7 = 14π cm

(b) A = π × 7² = π × 49 = 49π cm²

EEvaluate. Linear units (cm) for circumference; square units (cm²) for area. The numerical coefficients are different (14 vs 49) — they should be, because A = πr² grows faster than C = 2πr.

C = 14π cm; A = 49π cm²

02Worked Example — Circle (diameter given)

A circle has diameter 10 cm. Find (a) its circumference and (b) its area. Give exact answers in terms of π. (3 marks)

MMethod. For the circumference, C = πd is fastest when d is given. For the area, the formula needs the radius — halve d first to get r = 5.

CCalculate.

(a) C = πd = π × 10 = 10π cm

(b) r = d/2 = 5

A = π × 5² = π × 25 = 25π cm²

EEvaluate. Common wrong answer: A = π × 10² = 100π. That uses the diameter directly — four times too big. Always halve the diameter before substituting into A = πr².

C = 10π cm; A = 25π cm²

Sector & Arc — A Fraction of the Circle

A sector is a "pie slice" cut from a circle: two radii and a curved arc. The arc is just the curved part. Both depend on the angle θ at the centre.

The Fraction Idea

A full circle has angle 360° at its centre. A sector with angle θ takes up the fraction θ/360 of the whole circle. So:

Arc length = (θ/360) × full circumference = (θ/360) × 2πr

Sector area = (θ/360) × full area = (θ/360) × πr²

Arc Length Formula

Arc length = (θ/360) × 2πr

Step-by-step routine

Identify θ (sector angle, in degrees) and r (radius)

If you were given d, halve it to get r

Simplify the fraction θ/360 if possible (e.g. 72/360 = 1/5)

Multiply by 2πr, keeping π as a symbol

Quote the answer with linear units (cm) — it's a length, not an area

Sanity Check & Sector Perimeter

Sanity check: an arc length cannot exceed the full circumference 2πr. If your answer is bigger than 2πr, you've made an error.

Sector perimeter (sometimes asked separately): the boundary of a sector is the curved arc plus the two straight radii, so:

Sector perimeter = arc + 2r = (θ/360) × 2πr + 2r

For example, a sector with r = 10 and θ = 72° has arc = 4π, so its full perimeter is (4π + 20) cm.

Sector Area Formula

Sector area = (θ/360) × πr²

Special cases — memorise the fractions

Semicircle (θ = 180°): fraction = 180/360 = ½. Half the full area; half the circumference for the arc.

Quarter circle (θ = 90°): fraction = 90/360 = ¼.

Sixth (θ = 60°): fraction = 1/6.

Fifth (θ = 72°): fraction = 1/5.

Third (θ = 120°): fraction = 1/3.

Don't Forget the Fraction

A common error is to write the sector area as just πr² and forget the (θ/360) fraction — that gives the area of the whole circle. Always include the fraction.

03Worked Example — Sector (arc + area)

A sector has radius 10 cm and angle 72°. Find (a) the arc length, (b) the sector area. Give exact answers in terms of π. (4 marks)

MMethod. Arc length = (θ/360) × 2πr; sector area = (θ/360) × πr². Substitute θ = 72°, r = 10. The fraction 72/360 simplifies to 1/5.

CCalculate.

(a) Arc = (72/360) × 2π × 10 = (1/5) × 20π = 4π cm

(b) Area = (72/360) × π × 10² = (1/5) × 100π = 20π cm²

EEvaluate. Cross-check: full circumference 2π(10) = 20π; one-fifth is 4π ✓. Full area π(10²) = 100π; one-fifth is 20π ✓. Linear units for the arc; square units for the area.

Arc = 4π cm; Sector area = 20π cm²

Semicircles & Quarter Circles

You will see these constantly inside composite-shape questions. Memorise the fractions:

ShapeAngleFractionArc lengthArea

Full circle360°12πrπr²

Semicircle180°½πr½πr²

Quarter circle90°¼½πr¼πr²

Two Useful Forms for a Semicircle

The semicircle arc is sometimes written ½ × 2πr = πr, and sometimes ½ × πd = πd/2. Both are equal — pick whichever uses the value you've been given (radius or diameter) without halving.

Composite Shapes — Polygon + Circle Parts

A common exam shape combines a rectangle (or triangle) with a semicircle, quarter circle, or full sector. Treat each part separately and combine.

Routine for area

Identify the polygon part(s) and the circular part(s)

Find each piece's area using its own formula

Add (or subtract, if there is a cut-out) to get the total

Combine like terms — keep the polygon area numerical and the circular area in terms of π

Routine for perimeter

Walk around the outside boundary; do not include any internal split lines

Add up each straight side and each curved arc

Express the answer as (numerical part) + (π part) cm if both occur

Common Trap — Internal Boundary

Where a semicircle meets a rectangle, the diameter of the semicircle is not part of the perimeter — it is inside the shape. Only count the curved arc plus the three open rectangle sides.

04Worked Example — Composite (rectangle + semicircle)

The shape below is made of a rectangle 8 cm wide and 6 cm tall, with a semicircle on top whose diameter is 8 cm. Find (a) the perimeter and (b) the area. Give exact answers in terms of π. (5 marks)

MMethod. Treat the shape as rectangle + semicircle. The semicircle has diameter 8, so radius 4. The diameter line (where the rectangle meets the semicircle) is internal — it is not on the perimeter.

CCalculate.

(a) Perimeter = bottom + left + right + semicircle arc

= 8 + 6 + 6 + (½ × π × 8) = 20 + 4π

(b) Area = rectangle + semicircle

= (8 × 6) + (½ × π × 4²) = 48 + 8π

EEvaluate. Perimeter has linear units (cm), area has square units (cm²). Each answer has a numerical part and a π part — that is normal for a composite. Do not try to combine 20 with 4π into a single number; leave them separate.

Perimeter = (20 + 4π) cm; Area = (48 + 8π) cm²

Annulus — The "Ring" Shape

An annulus is the region between two concentric circles. Its area is the big circle's area minus the small circle's area.

Annulus area = πR² − πr² = π(R² − r²)

Critical Algebra Trap

The shaded area is π(R² − r²), not π(R − r)². Squaring R − r is wrong: (R − r)² = R² − 2Rr + r², not R² − r². Compute R² and r² separately, then subtract.

Difference of Squares

R² − r² = (R − r)(R + r). For R = 8 and r = 5, this gives (3)(13) = 39 — much faster than 64 − 25. Use whichever you find easier; both are correct.

05Worked Example — Annulus

The diagram shows two concentric circles. The outer circle has radius 8 cm and the inner circle has radius 5 cm. Find the shaded area. Give your answer in terms of π. (3 marks)

MMethod. Shaded area = (outer circle area) − (inner circle area) = πR² − πr² = π(R² − r²). Substitute R = 8, r = 5.

CCalculate.

Outer area = π × 8² = 64π

Inner area = π × 5² = 25π

Shaded = 64π − 25π = 39π cm²

EEvaluate. Cross-check using difference of squares: π(8 − 5)(8 + 5) = π × 3 × 13 = 39π ✓. Common wrong answer: π(8 − 5)² = 9π — that comes from mistakenly squaring (R − r) instead of subtracting R² and r² separately.

Shaded area = 39π cm²

06Worked Example — Reverse (find r from arc length)

A sector has angle 60° and arc length 6π cm. Find the radius of the sector. (3 marks)

MMethod. Use the arc-length formula in reverse. Arc = (θ/360) × 2πr = 6π. Substitute θ = 60° and solve the equation for r. (Cross-references 2.25 — Forming & Solving Equations.)

CCalculate.

(60/360) × 2πr = 6π

(1/6) × 2πr = 6π

πr/3 = 6π

Divide both sides by π: r/3 = 6

Multiply by 3: r = 18 cm

EEvaluate. Sense-check: with r = 18 and θ = 60°, arc = (1/6) × 2π × 18 = (1/6) × 36π = 6π ✓. Linear units (cm) — this is a length, not an area.

r = 18 cm

Same Technique — Finding the Sector Angle

Spec G18 explicitly mentions "arc lengths, angles and areas of sectors". The same rearrangement finds θ when r and the arc are given.

Example: a sector has radius 12 cm and arc length 2π cm. Find θ.

(θ/360) × 2π × 12 = 2π

24πθ/360 = 2π → divide by π → 24θ/360 = 2

θ/15 = 2 → θ = 30°

The same approach works for sector area: set (θ/360) × πr² equal to the given area and solve for θ or r.

Examiner Alerts

The same handful of errors lose marks on circle questions every series:

Diameter Used Where Radius Belongs

What students write: A = π × 10² = 100π when 10 was the diameter. What to write instead: halve d to get r = 5, then A = π × 5² = 25π. Always check whether the given length is a radius or a diameter.

Replacing π with 3.14 Too Soon

What students write: 2 × 3.14 × 7 = 43.96, then quote that as the answer. What to write instead: when the question says "in terms of π", keep π as a symbol throughout — the answer here is 14π. Mark schemes explicitly say "do not allow use of a numerical value for π" when an exact form is required.

Forgetting the (θ/360) Fraction

What students write: Sector area = πr². What to write instead: Sector area = (θ/360) × πr². Without the fraction, you have given the area of the whole circle, not just the sector.

(R − r)² Instead of R² − r² for an Annulus

What students write: π(R − r)² = π × 3² = 9π for R = 8, r = 5. What to write instead: π(R² − r²) = π(64 − 25) = 39π. Compute the squares separately and subtract; do not square the difference.

Including the Internal Diameter on a Composite Perimeter

What students write: Perimeter of (rectangle + semicircle) = three rectangle sides + diameter + arc. What to write instead: the diameter is internal — it joins the rectangle to the semicircle but is not on the outer boundary. Only count the three rectangle sides + the curved arc.

Lesson Checkpoint

You should now be able to:

Four Core Formulae (memorised)

Circumference: C = 2πr = πd

Area: A = πr²

Arc length: ℓ = (θ/360) × 2πr

Sector area: S = (θ/360) × πr²

Special Cases & Composites

Semicircle: ½ × πr² for area; πr (or πd/2) for arc

Quarter circle: ¼ × πr² for area; ½πr for arc

Composite (polygon + circular): sum each piece; do not include internal boundaries on the perimeter

Annulus: π(R² − r²) — never π(R − r)²

Exam Discipline

Halve diameters before using A = πr²

Keep π as a symbol when "in terms of π" is asked

Linear units (cm) for arcs and circumferences; square units (cm²) for areas

Compose answers as (numerical) + (π part) for composite shapes

Up Next

Lesson 4.14 covers 3D Mensuration: surface area and volume of prisms, cylinders, pyramids, cones and spheres — extending the area formulae here into three dimensions.