To find the circumcenter, we need equations for the perpendicular bisectors of at least two sides. However, first you need to draw an equilateral triangle.
Example Solution:
Practice makes perfect
Drawing our triangle
We will start by drawing an arbitrary equilateral triangle, which is a triangle with congruent sides and angles. To make it easier for us, we will place one of the sides on the x-axis. To make our triangle, we will draw two congruent sides with an included angle of 60^(∘). We will start with the base of the triangle.
Next, we will use a protractor to measure a 60^(∘) angle with B as our midpoint.
With our ruler, we measure a 6 cm side along B and the blue marking.
Finally, by connecting A and C, we get an equilateral triangle with a side of 6.
To find the circumcenter, we need the perpendicular bisectors of at least two sides.
Finding perpendicular bisectors
By the Slopes of Perpendicular Lines Theorem, we know that horizontal and vertical lines are perpendicular. Since BC is a horizontal line, the perpendicular line will be vertical. Also, from the diagram, we can identify the midpoint of BC as (3,0). Therefore, our equation for the perpendicular bisector to BC is x=3.
To find the perpendicular bisector to AB, we have to find the side's midpoint. Since we do not the coordinates for A, We will determine the midpoint with a ruler. We know AB=6 cm so we can find the midpoint by going 3 cm along AB from either endpoint.
When we have the midpoint, we will use a protractor to determine where the perpendicular bisector will run.
Finally, using a ruler, we can draw our perpendicular bisector to AB.
Note that the perpendicular bisector to the triangle's third side will also intersect at the same point as the other two. We could find it as well, but it's enough finding two of them.
Finding the circumcenter
Where the perpendicular bisectors intersect, we find the triangle's circumcenter.
Note that the perpendicular bisector to the triangle's third side will also intersect at the same point as the other two. We could find it as well, but it's enough finding two of them.
Constructing the circumscribed circle
The circumcenter is equidistant to all of the triangle's vertices. Therefore, by using a compass and setting it's radius to the distance between the circumcenter and an arbitrary vertex on the triangle, we can draw a circle that passes through all three vertices.
