Circle and Inscribed Regular Polygon Relations

Quarter Wit, Quarter WisdomAs promised last week, let’s figure out the relations between the sides of various inscribed regular polygons and the radius of the circle.

We will start with the simplest regular polygon – an equilateral triangle. We will use what we already know about triangles to arrive at the required relations.

Look at the figure given below. AB, BC and AC are sides (of length ‘a’) of the equilateral triangle. OA, OB and OC are radii (of length ‘r’) of the circle.

The interior angles of an equilateral triangle are 60 degrees each. Therefore, angle OBD is 30 degrees (since ABC is an equilateral triangle, BO will bisect angle ABD). So, triangle BOD is a 30-60-90 triangle.

As discussed in your geometry book, the ratio of sides in a 30-60-90 triangle is 1:?3:2 therefore, a/2 : r = ?3:2 or a:r = ?3:1

Side of the triangle = sqrt(3) * Radius of the circle

You don’t have to learn up this result. You can derive it if needed. Note that you can derive it using many other methods. Another method that easily comes to mind is using the altitude AD. Altitude AD of an equilateral triangle is given by (sqrt(3)/2)*a. The circum center is at a distance 2/3rd of the altitude so AO (radius) = (2/3)* (sqrt(3)/2)*a = a/sqrt(3)

Or side of the triangle = sqrt(3) * radius of the circle

Let’s look at a square now.

AB is the side of the square and AO and BO are the radii of the circle. Each interior angle of a square is 90 degrees so half of that angle will be 45 degrees. Therefore, ABO is a 45-45-90 triangle. We know that the ratio of sides in a 45-45-90 triangle is 1:1:sqrt(2).

r:a = 1: sqrt(2)

Side of the square = sqrt(2)*Radius of the circle

Again, no need to learn up the result. Also, there are many methods of arriving at the relation. Another one is using the diagonal of the square. The diagonal of a square is sqrt(2) times the side of the square. The radius of the circle is half the diagonal. So the side is the square is sqrt(2)*radius of the circle.

The case of a pentagon is more complicated since it needs the working knowledge of trigonometry which is beyond GMAT scope so we will not delve into it.

We will look at a hexagon though.

Notice that the interior angle of a regular hexagon is 120 degrees so half of that will be 60 degrees. Therefore, both angles OAB and OBA will be 60 degrees each. This means that triangle OAB is an equilateral triangle with all angles 60 degrees and all sides equal. Hence,

Side of the regular hexagon = Radius of the circle.

The higher order regular polygons and more complicated and we will not take them up. We will discuss a circle inscribed in a polygon next week.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!