Polygons, Area, and Circles
Pillar: SHAPE — "Know the shape, know the formula."
Quadrilaterals
A quadrilateral is a four-sided polygon. The six types you need form a hierarchy — each inherits properties from its parent:
| Shape | Defining property | Angles | Diagonals |
|---|---|---|---|
| Parallelogram | Both pairs of opposite sides parallel | Opposite angles equal | Bisect each other |
| Rectangle | Parallelogram with four right angles | All \(90^\circ\) | Equal length, bisect each other |
| Rhombus | Parallelogram with four equal sides | Opposite angles equal | Perpendicular, bisect each other |
| Square | Rectangle + Rhombus | All \(90^\circ\) | Equal, perpendicular, bisect each other |
| Trapezoid | Exactly one pair of parallel sides | Varies | Varies |
| Kite | Two pairs of consecutive equal sides | One pair of opposite angles equal | Perpendicular (one bisected) |
The hierarchy: every square is a rectangle and a rhombus. Every rectangle and rhombus is a parallelogram. Knowing which type you have tells you which formulas and properties apply.
Interior Angle Sum
For any convex polygon with \(n\) sides, the sum of interior angles is:
Why? Pick any vertex and draw diagonals to all non-adjacent vertices. This divides the polygon into \(n - 2\) triangles. Each triangle contributes \(180^\circ\), and together they cover every interior angle exactly once.
| Polygon | \(n\) | Angle sum |
|---|---|---|
| Triangle | 3 | \(180^\circ\) |
| Quadrilateral | 4 | \(360^\circ\) |
| Pentagon | 5 | \(540^\circ\) |
| Hexagon | 6 | \(720^\circ\) |
| \(n\)-gon | \(n\) | \((n-2) \cdot 180^\circ\) |
For a regular polygon (all sides and angles equal), each interior angle is:
A regular hexagon has interior angles of \(\frac{4 \cdot 180}{6} = 120^\circ\). A regular octagon: \(\frac{6 \cdot 180}{8} = 135^\circ\).
The exterior angle at each vertex of a convex polygon is the supplement of the interior angle. Exterior angles always sum to \(360^\circ\) regardless of \(n\). For a regular polygon, each exterior angle is \(360^\circ / n\).
Area Formulas: Triangles
Three ways to compute the area of a triangle, each suited to different inputs.
Base-Height Formula
where \(b\) is any side (the base) and \(h\) is the perpendicular distance from the opposite vertex to that base (the height). This is the most common formula, but it requires knowing the height.
Heron's Formula
When you know all three side lengths \(a\), \(b\), \(c\) but not the height:
Example. Triangle with sides \(5\), \(6\), \(7\).
double heronArea(double a, double b, double c) {
double s = (a + b + c) / 2.0;
return sqrt(s * (s - a) * (s - b) * (s - c));
}
Heron's formula is the go-to in CP when coordinates are not given — just side lengths.
Two-Sides-and-Included-Angle Formula
When you know two sides \(a\), \(b\) and the angle \(\theta\) between them:
This connects directly to trigonometry (Lesson 4) and to the cross product in coordinate geometry (Lesson 3).
Area Formulas: Quadrilaterals
| Shape | Area formula | Notes |
|---|---|---|
| Rectangle | \(A = \ell w\) | length \(\times\) width |
| Parallelogram | \(A = b h\) | base \(\times\) perpendicular height |
| Rhombus | \(A = \frac{1}{2} d_1 d_2\) | product of diagonals \(\div 2\) |
| Trapezoid | \(A = \frac{1}{2}(b_1 + b_2) h\) | average of parallel sides \(\times\) height |
| Kite | \(A = \frac{1}{2} d_1 d_2\) | same as rhombus: half the diagonal product |
Worked example. A trapezoid has parallel sides \(8\) and \(14\), and height \(5\).
Every quadrilateral area formula reduces to "base times height" in some form. The parallelogram formula is literal. The trapezoid averages the two bases. The rhombus and kite use the diagonals as implicit bases and heights (because the diagonals are perpendicular).
Regular Polygons
A regular polygon has all sides equal and all interior angles equal. Two measurements define its geometry:
- Side length \(s\)
- Apothem \(a\) — the perpendicular distance from the center to the midpoint of a side
The area of any regular \(n\)-gon is:
The apothem relates to the side length via:
So the area in terms of side length alone:
Example. Regular hexagon with side length \(6\).
As \(n\) increases, a regular polygon approaches a circle. This is both a historical insight (Archimedes approximated \(\pi\) this way) and a useful sanity check: for large \(n\), \(A \approx \pi r^2\) where \(r\) is the circumradius.
Circles: Definitions
A circle is the set of all points at distance \(r\) (the radius) from a center point.
| Term | Definition |
|---|---|
| Radius (\(r\)) | Distance from center to any point on the circle |
| Diameter (\(d\)) | \(d = 2r\); longest chord, passes through center |
| Chord | Line segment with both endpoints on the circle |
| Arc | Portion of the circumference between two points |
| Sector | "Pie slice" — region bounded by two radii and an arc |
| Segment | Region between a chord and its arc |
| Tangent | Line touching the circle at exactly one point |
| Secant | Line crossing the circle at two points |
A tangent line is always perpendicular to the radius at the point of tangency. This single fact solves a surprising number of problems.
Circle Formulas
Circumference (perimeter):
Area:
These two formulas plus the angle-fraction idea below cover all basic circle computations.
Arc Length and Sector Area
An arc that subtends a central angle \(\theta\) (in degrees) spans a fraction \(\theta / 360\) of the full circle.
Arc length:
Sector area:
In radians, these simplify cleanly: arc length \(= r\theta\), sector area \(= \frac{1}{2}r^2\theta\).
Example. Circle with \(r = 8\), central angle \(60^\circ\).
double arcLength(double r, double angleDeg) {
return 2.0 * M_PI * r * angleDeg / 360.0;
}
double sectorArea(double r, double angleDeg) {
return M_PI * r * r * angleDeg / 360.0;
}
Inscribed Angle Theorem
An inscribed angle is formed by two chords that meet at a point on the circle. A central angle is formed by two radii.
Theorem: An inscribed angle is exactly half the central angle that subtends the same arc.
Corollaries:
- All inscribed angles that subtend the same arc are equal.
- An inscribed angle that subtends a semicircle (diameter) is \(90^\circ\). This is Thales' theorem.
- Opposite angles of a cyclic quadrilateral (inscribed in a circle) sum to \(180^\circ\).
Example. A central angle is \(100^\circ\). An inscribed angle subtending the same arc is \(50^\circ\). If you see a triangle inscribed in a circle with the hypotenuse as the diameter, the angle opposite the diameter is always \(90^\circ\) — that's Thales' theorem in action.
Power of a Point (Preview)
The power of a point theorem unifies several circle theorems. For a point \(P\) and a circle:
- If two chords through \(P\) intersect the circle at \((A, B)\) and \((C, D)\): \(PA \cdot PB = PC \cdot PD\).
- If \(P\) is outside the circle and a tangent from \(P\) touches at \(T\), and a secant through \(P\) hits the circle at \(A\) and \(B\): \(PT^2 = PA \cdot PB\).
The quantity \(PA \cdot PB\) is constant for all lines through \(P\) — that constant is the power of the point. This shows up in computational geometry when testing point-circle relationships.
Worked Examples
Example 1: Interior Angles of a Regular Polygon
A regular polygon has interior angles of \(140^\circ\). How many sides does it have?
Each interior angle of a regular \(n\)-gon is \(\frac{(n-2) \cdot 180}{n}\). Set equal to \(140\):
It's a regular nonagon (9 sides).
Example 2: Heron's Formula
Find the area of a triangle with sides \(13\), \(14\), \(15\).
Example 3: Circle Segment Area
Find the area of the segment cut off by a chord that subtends a central angle of \(90^\circ\) in a circle of radius \(10\).
Segment area \(=\) sector area \(-\) triangle area.
The triangle formed by the two radii and the chord is a right isosceles triangle with legs \(10\):
Practice Problems
1. Find the sum of interior angles of a 12-sided polygon.
2. A parallelogram has base \(9\) and height \(6\). Find its area.
3. Use Heron's formula to find the area of a triangle with sides \(10\), \(17\), \(21\).
4. A circle has area \(50\pi\). Find its circumference.
5. A sector of a circle with radius \(12\) has arc length \(4\pi\). Find the central angle and the sector area.
6. An inscribed angle in a circle subtends an arc of \(130^\circ\). What is the inscribed angle?
Answers
1. \((12 - 2) \cdot 180 = 10 \cdot 180 = 1800^\circ\).
2. \(A = 9 \cdot 6 = 54\).
3. \(s = (10 + 17 + 21)/2 = 24\). \(A = \sqrt{24 \cdot 14 \cdot 7 \cdot 3} = \sqrt{7056} = 84\).
4. \(\pi r^2 = 50\pi\), so \(r = 5\sqrt{2}\). \(C = 2\pi \cdot 5\sqrt{2} = 10\pi\sqrt{2} \approx 44.43\).
5. Arc length \(= \frac{\theta}{360} \cdot 2\pi \cdot 12 = 4\pi\). So \(\frac{\theta}{360} \cdot 24\pi = 4\pi\), giving \(\theta = 60^\circ\). Sector area \(= \frac{60}{360} \cdot \pi \cdot 144 = 24\pi\).
6. The inscribed angle is half the central angle. The central angle subtending the same arc is \(130^\circ\). So the inscribed angle is \(65^\circ\).
Quick Recap
- Quadrilateral hierarchy: parallelogram \(\supset\) rectangle, rhombus \(\supset\) square. Trapezoid and kite are separate.
- Interior angle sum of an \(n\)-gon: \((n-2) \cdot 180^\circ\). Regular polygon: each angle \(= (n-2) \cdot 180 / n\).
- Triangle area: \(\frac{1}{2}bh\) (base-height), Heron's \(\sqrt{s(s-a)(s-b)(s-c)}\) (three sides), \(\frac{1}{2}ab\sin\theta\) (two sides + angle).
- Circle: \(C = 2\pi r\), \(A = \pi r^2\). Arc and sector scale by \(\theta / 360\).
- Inscribed angle \(= \frac{1}{2} \times\) central angle. Angle in a semicircle is \(90^\circ\) (Thales).
- Regular polygon area \(= \frac{1}{2} \cdot \text{perimeter} \cdot \text{apothem}\).