• ## Top Posts

 quadratic equation c… on Modeling Data With Quadratic… Lemuel on Quadratic Functions and Their… Arizona Bayfield on How To Simplify Rational … Suemac on Proving Lines Parallel with Tr… Mr. Pi on Subsets of Real Numbers

## How to Find the Area of Regular Polygons

Finding the area of regular polygons is the topic of this geometry math video.

First, the video discusses the parts of a regular polygon: apothem and radius. The center of the polygon is the same as the center of the circumscribed circle. The radius of the polygon is the distance from the center of the polygon to its vertex. The number of radii is determined by the number of sides. The apothem it the perpendicular distance from the center of the polygon to a side.

The first example models how to find the different angles of a regular polygon formed by the radius and the apothem. First, you divide the number in interior angles into 360 degrees. Once you have that you can figure out the other two angles quite easily.

Before applying the the area formula of a regular polygon, the video reviews the formula:

$A = \frac{1}{2}ap$

It is good to note that a = length of the apothem and p = perimeter of the polygon. The perimeter may not be calculated. In that case you multiply the number of sides by the length of each side.

This example is similar to the video, but it is different:

What is the area of a polygon with sixteen 36 in. sides and an apothem of $18\sqrt{3} in.$

$A = \frac{1}{2}ap$

$A = \frac{1}{2}(18 \sqrt{3}(16*36))$

$A = 5184 \sqrt{3} in^2 \approx 8979.0 in^2$

Questions for the comment section:

Comparing example 2T on the video and the problem above, how are the problems alike? How are the problems the same?

Example 3T is finding the area of a regular polygon, but you need to find the length of the apothem using the 30-60-90 triangle relationship.

## Findng Inverses of Formulas and Compositions of Inverse Functions

In this video math lesson, working with inverse functions is discussed. It is good to remember, when finding the the inverse of a formula, DO NOT SWITCH THE VARIABLES. The first example models finding the inverse of an existing formula:
$d = 16t^2$
$16t^2 = d$ Symmetric Property of Equality
$t^2 = \frac{d}{16}$ Division Property of Equality
$\sqrt {t^2} = \sqrt{ \frac{d}{16}}$ Inverse of Square is Square Root
$t^ = \frac{ \sqrt{d}}{4}$ Simplify Fraction to Simplest Radical Form

I know some readers may not be able to follow the above problem. The same problem, but with the example in the video, the auditory learner can benefit too.

The second example models finding the composition of a function and its inverse. This can be written as $f^{-1}(f(x))$ or $f(f^{-1}(x))$. In both cases are equal to the value of x. Performing the composition of a function and its inverse gives the value you started with. You will see in the video, how simple this process is.

In this post, I have embedded an Algebra 2 Video Math Lesson about the function $f(x) = \sqrt {2x+2}$. Before considering the function, the video defines $f^{-1}$ as the inverse of f or as f inverse.
1. Find the domain and range of $f(x) = \sqrt {2x+2}$
2. Find $f^{-1}$
3. Find the domain and range of $f^{-1}$
4. Determine if $f^{-1}$ is a function