Understanding Polygons and Their Angles

Before diving into the formula, let’s clarify what polygons are and the types of angles we’ll be dealing with.

What is a Polygon?

A polygon is a closed figure formed by straight line segments. These segments are called sides, and where two sides meet, they form a vertex (plural: vertices). Here are some examples of polygons:

  • Triangle: A polygon with 3 sides and 3 vertices.
  • Quadrilateral: A polygon with 4 sides and 4 vertices.
  • Pentagon: A polygon with 5 sides and 5 vertices.
  • Hexagon: A polygon with 6 sides and 6 vertices.
  • Heptagon: A polygon with 7 sides and 7 vertices.
  • Octagon: A polygon with 8 sides and 8 vertices.
  • Nonagon: A polygon with 9 sides and 9 vertices.
  • Decagon: A polygon with 10 sides and 10 vertices.

Types of Angles in a Polygon

There are two main types of angles we’ll focus on in polygons:

1. Interior Angles

Interior angles are the angles formed inside the polygon. They are the angles formed by two adjacent sides of the polygon.

2. Exterior Angles

Exterior angles are the angles formed outside the polygon when one side is extended. They are formed by a side of the polygon and its extension.

Finding the Sum of Interior Angles

The sum of interior angles in a polygon is determined by the number of sides it has. Here’s the formula:

Sum of Interior Angles = (n – 2) * 180°

Where ‘n’ represents the number of sides in the polygon.

Explanation of the Formula

This formula is based on the idea of dividing a polygon into triangles. Here’s how it works:

  1. Divide the polygon into triangles: You can always divide a polygon into triangles by drawing diagonals from one vertex to all other non-adjacent vertices. The number of triangles you can form is always two less than the number of sides (n – 2). For example, a quadrilateral can be divided into two triangles, a pentagon into three triangles, and so on.

  2. Sum of angles in a triangle: The sum of interior angles in any triangle is always 180°. This is a fundamental fact in geometry.

  3. Combining the triangles: Since each triangle contributes 180° to the total sum, and you have (n – 2) triangles, the total sum of interior angles is (n – 2) * 180°.

Examples

Let’s illustrate this with some examples:

Example 1: Sum of angles in a quadrilateral

A quadrilateral has 4 sides (n = 4). Using the formula:

Sum of Interior Angles = (4 – 2) * 180° = 2 * 180° = 360°

Therefore, the sum of interior angles in a quadrilateral is 360°.

Example 2: Sum of angles in a hexagon

A hexagon has 6 sides (n = 6). Using the formula:

Sum of Interior Angles = (6 – 2) * 180° = 4 * 180° = 720°

Therefore, the sum of interior angles in a hexagon is 720°.

Finding a Single Interior Angle

If you need to find the measure of a single interior angle in a regular polygon (a polygon where all sides and angles are equal), you can use the following formula:

Single Interior Angle = (n – 2) * 180° / n

Where ‘n’ is the number of sides.

Example: Finding an angle in a regular pentagon

A regular pentagon has 5 sides (n = 5). Using the formula:

Single Interior Angle = (5 – 2) * 180° / 5 = 3 * 180° / 5 = 108°

Therefore, each interior angle in a regular pentagon measures 108°.

Conclusion

Understanding how to calculate the sum of interior angles in a polygon is essential for many geometric problems. This knowledge allows you to analyze and solve problems related to shapes, angles, and their relationships within polygons.

3. CK-12 – Interior and Exterior Angles of Polygons

Citations

  1. 1. Khan Academy – Angles in Polygons
  2. 2. Math is Fun – Angles in Polygons

Related

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H + HO2 → O2 + H2 k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O2 k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) H + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-5 s^-1) φ

Table 1 Reactions, rate constants and activation energies used in the model* No. Reaction kopt (M⁻¹ s⁻¹) 1 OH + H₂ → H + H₂O 3.74 x 10⁷ 2 OH + HO₂ → HO₂ + OH⁻ 5 x 10⁹ 3 OH + H₂O₂ → HO₂ + H₂O 3.8 x 10⁷ 4 OH + O₂ → O₂ + OH 9.96 x 10⁹ 5 OH + HO₂ → O₂ + H₂O 7.1 x 10⁹ 6 OH + OH → H₂O₂ 5.3 x 10⁹ 7 OH + e⁻aq → OH⁻ 3 x 10¹⁰ 8 H + O₂ → HO₂ 2.0 x 10¹⁰ 9 H + HO₂ → H₂O₂ 2.0 x 10¹⁰ 10 H + H₂O₂ → OH + H₂O 3.44 x 10⁷ 11 H + OH → H₂O 1.4 x 10¹⁰ 12 H + H → H₂ 1.94 x 10¹⁰ 13 e⁻aq + O₂ → O₂⁻ 1.9 x 10¹⁰ 14 e⁻aq + O₂ → HO₂⁻ + OH⁻ 1.3 x 10¹⁰ 15 e⁻aq + HO₂ 2.0 x 10¹⁰ 16 e⁻aq + H₂O₂ 1.1 x 10¹⁰ 17 e⁻aq + HO₂ → OH + OH⁻ 1.3 x 10¹⁰ 18 e⁻aq + H⁺ → H 2.3 x 10¹⁰ 19 e⁻aq + e⁻aq → H₂ + OH⁻ + OH⁻ 2.5 x 10⁹ 20 HO₂ + O₂ → O₂ + HO₂ 1.3 x 10⁹ 21 HO₂ + HO₂ → O₂ + H₂O₂ 8.3 x 10⁵ 22 HO₂ + HO₂ → O₂ + OH + H₂O 3.7 23 HO₂ + HO₂ → O₂ + O₂ + OH + H₂O 7 x 10⁵ s⁻¹ 24 H⁺ + O₂⁻ → HO₂ 4.5 x 10¹⁰ 25 H⁺ + O₂⁻ → O₂ 2.0 x 10¹⁰ 26 H⁺ + OH⁻ 1.4 x 10¹¹ 27 H⁺ + HO₂⁻ 2 x 10¹⁰ 28 H₂O₂ → HO₂ + H⁺ + OH⁻ 2.5 x 10⁻⁵ s⁻¹ 29 H₂O₂ → H⁺ + OH⁻ 1.4 x 10⁻⁷ s⁻¹ 30 O₂ + O₂ → O₂ + HO₂ + OH⁻ 0.3 31 O₂ + H₂O₂ → O₂ + OH + OH 16 32

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H2O + O → 2 OH k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) OH + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-8 s^-1) φ