UNIT circle triangle problem

Question

An equilateral triangle is inscribed in the unit circle. find the exact perimeter of the equilateral triangle. 

Answer 1

There are a few different ways to do this, but here's my favorite:

This triangle being inscribed in the circle means its vertices are on the circumference. Let's use that in a clever way. Place one vertex at the point (1,0) on our unit circle.

Now, the x-axis is both an angle bisector for that vertex and a perpendicular bisector.

This tells us several things, including that the top half of the angle is 30 degrees. We're going to use that with one of our theorems to figure out where our other two vertices are.

The Central Angle Theorem states that the measure of inscribed angle (∠APB) is always half the measure of the central angle ∠AOB.

If we set one endpoint of the arc to be at (-1,0) and the other to be on our triangle, with point P as our known (1,0), we'll find the central angle between the negative x-axis and our triangle is 60 degrees.

What's 60 degrees before 180 degrees? 120 degrees. That point is at (-1/2, sqrt(3)/2).

If we mirror this for the lower half of our triangle, we'll find the third vertex. at 240 degrees. Then you'll calculate the distance between any two vertices (I think finding the length of the side between vertex 2 and vertex 3 to be easiest here), because it's equilateral you don't need to calculate it thrice. You can if you want to check your points, though.

And finally, use that side length to find the perimeter

Answer 2

I encourage you do make the appropriate drawing. In a unit circle, inscribe an equilateral triangle.

From the center, draw radii to two of the vertices, giving you a triangle with a central angle measuring 120^o.

Draw the perpendicular bisector to the equilateral's side (which will also bisect that central angle).

Now you have two 30-60-90 triangles.

Concentrate on one of these; the hypotenuse is 1 (radius of circumscribing circle).

The shorter leg has length ½ (because it's opposite a 30^o angle & equals half the hypotenuse).

The other leg measures therefore ½√3.

Therefore the equilateral triangle's side is of length √3 and so the perimeter of the equilateral triangle is 3√3.