We covered a *lot* in Geometry this year, and overall, I found trigonometry to be one of the easiest units. Once you’re familiar with the mechanics, it all comes down to calculation. However, its implementation can be a bit more complicated, like in finding the surface area of a 2D shape with more than four sides, but the process is actually pretty cool.

I mentioned last time that with Geometry, everything seems to tie back to one another. And surprisingly, in finding the area of a *polygon *(of which a circle is not), a circle is the most necessary element. This comes down to **circumscribed circles. **This is a circle that completely surrounds all vertexes of a polygon. For this lesson, we’ll only consider **regular polygons, **meaning that all sides and angles are equal (creating equal arcs, and thus equal *central angles, *which are necessary to finding area). This also means that ALL vertexes are touching the circle, and the center point of the polygon is equidistant from all vertexes and points on the circle.

The particular shape I’ll be focusing on is a *hexagon, *especially since we dealt with a lot of problems involving these in Project Lead the Way this year, which were daunting and confusing from the start (until we actually learned how to *properly* find the area in Geometry).

Take this, an inscribed (the opposite of circumscribed) hexagon:

Our goal is to find the surface area of the hexagon. To do that, we’ll need a bit more information. Let’s say the **radius = 7 inches. **Now that the circle and hexagon are layered on top of each other, we can see that the radius of the circle is the same as the distance from the center point to any one the vertexes. This is why having a circle to reference is so helpful. To visualize this, let’s divide up the hexagon with lines running (approximately) from each vertex to the center point.

Now, our shape is divided up into six triangles, familiar territory for trigonometry. Since this is a regular polygon and all angles are equal, we can use the principle involving central angles that we went over in the previous lesson to determine that each central angle is worth 60 degrees.

Since we have an isosceles triangle, we can easily split one of these triangles into two equal *right triangles *with an __altitude__ (a line drawn perpendicular to a side from a vertex). This line would be considered an angle bisector, and so the two angles created would both be equal to 30 degrees. Here’s what the overall shape would look like, along with one individual triangle, which is what we’ll work with from here on out:

(not drawn to scale)

There is a formula for finding the area of regular polygons, and it is:

**1/2 * apothem * perimeter**

In this case, the variable *x* represents the apothem, the line we drew in on the left figure. So, let’s use simple trig to find the value for x.

**cos(30) = x / 7**

If you could visualize this vertically, then we can cross multiply using a calculator, and find that the length of the apothem is **6.06 inches **(rounded to the nearest hundredth).

In the same way, let’s find the value of y, which will help us find the perimeter:

**sin(30) = y / 7**

**y = 3.50in**

In order to find the perimeter, we can first multiply the length of the corresponding leg of the triangle (which, remember, is 1/2 of the full side length) to find an entire side length. Then, we can multiply that value by the number of sides to find our perimeter. In this case, our perimeter is **42 inches.**

Now, we can substitute our values in for our variables:

**a = 1/2 * 6.06 * 42**

**a = 127.26in**

With these types of problems, be sure to follow the rounding procedures as instructed. For example, you may not want to round until the very end.

And, as always, problems won’t always be set up like this. For example, you might be given the length of the apothem initially and that’ll save you an extra step. Or the problem might be much more complicated and make you solve, for example, for the radius first before being able to proceed. Whatever the problem, you’ll want to solve for whichever variables you need through isolation (simple algebra).

These processes can also be applied to polygons of varying side numbers (like a dodecagon, for instance). With these, you’ll just have to pay attention to finding the right central angle.

Next, we’ll work with finding surface areas and volumes of *3D objects.*

Thanks so much for reading!

Gabe