Circles and Polygons formulas
At a glance
Chord: r² = d² + (c/2)² ∠ at centre = 2 × ∠ at circumference; semicircle → 90° Same segment: equal angles; cyclic quad: opposite ∠s = 180° Tangent ⟂ radius; PT = PT′ = √(d² − r²) Common tangents: direct √(d² − (r₁ − r₂)²), transverse √(d² − (r₁ + r₂)²) Apart 4, touching outside 3, cutting 2, touching inside 1, inside 0 PA × PB = PC × PD; PT² = PA × PB; tangent-chord ∠ = alternate segment ∠ Arc = θ/360 × 2πr; sector = θ/360 × πr² Interior sum = (n − 2) × 180°; exterior = 360°/n Diagonals = n(n − 3)/2 Rhombus, kite ½d₁d₂; trapezium ½(a + b)h; parallelogram bh Cyclic quad: √((s − a)(s − b)(s − c)(s − d)) Regular hexagon: (3√3/2)a², diagonals 2a and a√3
Chords and the radius
OM ⟂ AB ⟺ AM = MB r² = d² + (c/2)² (c = chord, d = distance from O)
Angle at the centre is twice the angle at the circumference
∠AOB = 2 × ∠APB (same arc AB) Angle in a semicircle = 90°
Shortcut: The converse is just as useful: in any right triangle, the hypotenuse is a diameter of its circumcircle. Circumradius = hypotenuse / 2.
Same segment and cyclic quadrilaterals
Same segment: ∠APB = ∠AQB Cyclic quadrilateral: ∠A + ∠C = ∠B + ∠D = 180° Exterior angle = interior opposite angle
Shortcut: Extend one side of a cyclic quadrilateral. The exterior angle you get equals the interior angle at the opposite corner, with no working needed.
Tangents
Tangent ⟂ radius at the point of contact PT = PT′ = √(d² − r²) ∠TPT′ + ∠TOT′ = 180°
direct (both circles on the same side): √(d² − (r₁ − r₂)²) transverse (crosses between them): √(d² − (r₁ + r₂)²) r₁ = 8, r₂ = 3, d = 13: direct √(169 − 25) = 12, transverse √(169 − 121) = 4√3
| Position | Distance between centres | Direct | Transverse | Total |
|---|---|---|---|---|
| one inside the other | d < r₁ − r₂ | 0 | 0 | 0 |
| touching inside | d = r₁ − r₂ | 1 | 0 | 1 |
| cutting at two points | r₁ − r₂ < d < r₁ + r₂ | 2 | 0 | 2 |
| touching outside | d = r₁ + r₂ | 2 | 1 | 3 |
| apart | d > r₁ + r₂ | 2 | 2 | 4 |
Concentric circles (d = 0) are the extreme case of one inside the other: no common tangent.
Secants, crossing chords and the alternate segment
Chords AB and CD cross at P inside: PA × PB = PC × PD Secants PAB and PCD from P outside: PA × PB = PC × PD Tangent PT and secant PAB: PT² = PA × PB Alternate segment: ∠(tangent, chord AB) at A = angle AB makes at any point in the other segment
Shortcut: A tangent at A and the chord AB make 70°. Then every point C on the arc away from that angle sees AB at 70° too, so ∠ACB = 70°. It is the 'same segment' rule with the tangent standing in for a chord.
Sectors and arcs
Arc length = θ/360 × 2πr Sector area = θ/360 × πr² = ½ × arc × r Sector perimeter = arc + 2r
Angles of a polygon
Sum of interior angles = (n − 2) × 180° Regular: each exterior = 360°/n, each interior = 180° − 360°/n
Shortcut: Work through the exterior angle. It is small and 360/n is easy. Interior 156° means exterior 24°, so n = 360/24 = 15.
Diagonals of a polygon
Diagonals = n(n − 3)/2 = nC2 − n From one vertex: n − 3, making n − 2 triangles
| Sides n | Name | Angle sum | Each interior | Each exterior | Diagonals |
|---|---|---|---|---|---|
| 3 | triangle | 180° | 60° | 120° | 0 |
| 4 | square | 360° | 90° | 90° | 2 |
| 5 | pentagon | 540° | 108° | 72° | 5 |
| 6 | hexagon | 720° | 120° | 60° | 9 |
| 7 | heptagon | 900° | 128⁴⁄₇° | 51³⁄₇° | 14 |
| 8 | octagon | 1080° | 135° | 45° | 20 |
| 9 | nonagon | 1260° | 140° | 40° | 27 |
| 10 | decagon | 1440° | 144° | 36° | 35 |
| 12 | dodecagon | 1800° | 150° | 30° | 54 |
Area = 6 × (√3/4)a² = (3√3/2)a² long diagonal (through the centre) = 2a; short diagonal = a√3 circumradius = a; inradius = (√3/2)a
Quadrilaterals and their areas
Parallelogram = base × height Rhombus = ½ × d₁ × d₂, side² = (d₁/2)² + (d₂/2)² Trapezium = ½ × (a + b) × h Square = side² = ½ × diagonal²
| Shape | Sides and angles | Diagonals | Area | Cyclic? |
|---|---|---|---|---|
| Parallelogram | opposite sides equal and parallel; opposite angles equal | bisect each other; d₁² + d₂² = 2(a² + b²) | base × height = ab sin θ | only if a rectangle |
| Rectangle | every angle 90° | bisect each other and are equal | l × b | always |
| Rhombus | all four sides equal | bisect each other at 90° | ½ d₁d₂ | only if a square |
| Square | all sides equal, every angle 90° | equal, and bisect each other at 90° | a² = ½ d² | always |
| Kite | two pairs of equal neighbouring sides | perpendicular; the axis bisects the other | ½ d₁d₂ | only if the angles between unequal sides are 90° |
| Trapezium | one pair of parallel sides a, b | equal only in an isosceles trapezium | ½ (a + b) × h | only if isosceles |
| Cyclic quadrilateral | opposite angles add to 180° | AC × BD = AB × CD + AD × BC (Ptolemy) | √((s − a)(s − b)(s − c)(s − d)) | yes, by definition |
s = (a + b + c + d)/2. The cyclic area formula (Brahmagupta) is Heron's formula with a fourth side.
Shortcut: Sides 7, 24, 15, 20 round a circle of diameter 25 (7-24-25 and 15-20-25 triangles on the diameter). s = 33, so area = √(26 × 9 × 18 × 13) = 234, which matches 84 + 150 from the two right triangles.
Formulas stick when you use them
The Circles and Polygons lesson has worked problems and quick checks for every formula here. Free to start.
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