Learn first: Quadrilaterals
A polygon is a closed plane figure made of line segments (sides) joined end to end. Triangles, quadrilaterals and pentagons are all special cases of polygons. Polygon problems rest on two main ideas: the angle sum (found by splitting the polygon into triangles) and splitting a regular polygon into equal triangles meeting at its centre. The second idea makes problems with inscribed and circumscribed circles easy. Questions about angles, the number of diagonals, radii and area on this topic appear in the exam almost every year.
A polygon with n vertices is called an n-gon; it has n sides and n interior angles. A segment joining two non-adjacent vertices is a diagonal.
A convex polygon lies entirely on one side of the line through any of its sides. Every interior angle of a convex polygon is less than 180°, and all its diagonals lie inside it. In the school course and in the exam, a polygon is assumed to be convex unless stated otherwise.
The perimeter P is the sum of the lengths of all the sides.
Draw all the diagonals of a convex n-gon from one vertex. There are n − 3 of them (no diagonal can go to the vertex itself or to its two neighbours), and they split the polygon into n − 2 triangles. The angles of each triangle add up to 180°, and together the angles of the triangles make up exactly the angles of the polygon. Therefore:
Interior angle sum S = 180°(n − 2)
Check: triangle 180°, quadrilateral 360°, pentagon 540°, hexagon 720°. Each extra vertex adds 180°.
In the reverse problem we find n from the sum: n = S/180° + 2. If S is not divisible by 180°, no such polygon exists.
The interior angles of a polygon add up to 1440°. How many sides does it have?
An exterior angle is the angle adjacent to an interior angle, i.e. the angle between one side and the extension of the neighbouring side. At each vertex the interior and exterior angles add up to 180°. Over all vertices: 180°n − 180°(n − 2) = 360°.
Exterior angle sum (one per vertex) 360° — independent of n
Think of a person walking round the polygon: at each vertex they turn through the exterior angle, and by the end they have turned exactly once, i.e. through 360°.
Number of diagonals. From each vertex there are n − 3 diagonals, n(n − 3) in total. But each diagonal is counted twice (from both ends), so we halve it:
Number of diagonals d = n(n − 3)/2
For example, a decagon has 10 · 7/2 = 35 diagonals. If there are 20 diagonals, then n(n − 3) = 40 → n = 8.
A regular polygon is a convex polygon whose sides are all equal and whose angles are all equal. A regular triangle is equilateral; a regular quadrilateral is a square. Note that a rhombus has equal sides but unequal angles, so it is not regular; a rectangle has equal angles but unequal sides.
Since the angles are equal, we divide the sum by n:
Interior angle α = 180°(n − 2)/n, exterior angle β = 360°/n
The quickest way to find n is via the exterior angle: interior 140° → exterior 180° − 140° = 40° → n = 360/40 = 9.
The exterior angle of a regular polygon must divide 360°. So, for example, there is no regular polygon with an interior angle of 130°: the exterior angle is 50°, and 360/50 is not a whole number.
Find the interior angle of a regular 12-gon.
Every regular polygon has a circumscribed circle (through the vertices) and an inscribed circle (touching the sides), with the same centre. Joining the centre to the vertices gives n equal isosceles triangles; each has angle 360°/n at the centre, legs R and base equal to the side a. The height of such a triangle is the inradius r (the apothem). The height splits it into two right triangles with angle 180°/n at the centre. Hence:
a = 2R sin(180°/n), r = R cos(180°/n), a = 2r tan(180°/n)
Special cases: triangle a = R√3, r = R/2; square a = R√2, r = a/2; hexagon `a = R`, r = a√3/2
In a hexagon a = R because the central angle is 60°, so all six triangles are equilateral.
By Pythagoras, R² = r² + (a/2)². So the area of the ring between the circumscribed and inscribed circles depends only on the side:
π(R² − r²) = π(a/2)²
An equilateral triangle is inscribed in a circle of radius 6. Find its side, its inradius and its area.
The circle inscribed in a regular octagon has radius r = 2 + √2. Find the area of the ring between the inscribed and circumscribed circles.
The polygon consists of n equal triangles, each of area ½ a r. Therefore:
S = ½ P · r (half the perimeter times the inradius)
S = ½ n R² sin(360°/n)
Regular hexagon: S = 6 · (a²√3/4) = 3√3a²/2
For example, a square inscribed in a circle of radius 5 has area ½ · 4 · 25 · sin 90° = 50 (or a = 5√2, a² = 50).
A regular hexagon has side 4. Find R, r and its area.