When Algebra and Geometry Meet

Putting the pieces together

In my last post, I introduced the idea of systematically solving problems by first finding the hidden equations in a problem, and then using algebra to solve those equations and find the answer. In this post, I will expand on that idea in a more complicated problem that relies on some geometry knowledge. Here we go…

A circle has a circumference of c feet and an area of a square feet. If c = 3a, what is the radius of the circle?

This problem seems simple at first. It is barely more than a line, and it doesn’t contain any particularly scary equations. However, solving it proves to be difficult. The problem is that you want to find the radius (let’s call it r), which is related to c, and also related to a, neither of which we know. All we know is the relationship between c and a, which is a good start, but not enough to solve the problem.

Finding the Equations

Let’s take stock of our variables. We have three of them: c, a, and r.

Remember, to solve algebraic equations, we need one unique equation for each variable we use. We have one equation already (c = 3a), so we need to find two more.

But where do they come from? It is very important to realize that you can use equations you’ve learned in school about geometry. This is why formulas are useful! They help us solve problems that would otherwise be unsolvable by giving us another equation.

Circumference, area, radius. Think… what two equations do we know that relate these two quantities. Let’s start with circumference. It equals pi times the diameter of the circle. In other words:

\(c = 2 \pi r\)

What about the area? Thinking back to geometry, we know that:

\(a = \pi r ^2\)

Solving the Equations

All right! Now at this point, if I were taking a test, I would breathe a sigh of relief. You may not realize it yet, but we’re almost done. We have 3 variables and 3 unique equations, which means that we can solve our equations and get r.

Let’s make a substitution. Since c = 3a, and c = 2πr , then by substitution, or transitive property:

\(3a = 2\pi r \)

Now, we have 2 equations that have only a and r. So, using the fact that a = πr^2, we can substitute πr^2 for a in the above equation. This gives us:

\(3\pi r^2 = 2\pi r \)

Almost there! Now we have one equation that only contains r. We can divide both sides by π and by r to cancel them out. This leaves us with:

\(3r = 2\)

Finally, we can divide both sides by 3 to solve for r:

\(\mathbf{r = 2/3}\)

We got it! If we want to check our work, we can plug in to solve for c and a:

\(c = 2\pi (2/3) = 4/3 *\pi \)

\(a = \pi (2/3)^2 = 4/9 *\pi \)

Thus, we can verify that indeed, c = 3a, confirming that our answer is correct.

Conclusion

Smart variable choice is key in getting the necessary equations to solve a problem. The answer you are looking for should either be one of the variables or easily calculated from them. Two of the variables here were given to us in an equation, and the choice of r as a third variable was natural due to its relation to both area and circumference.

Then, when it comes to finding equations, use both given information from the problem and topic-specific formulas that relate to the problem.

Additionally, keep in mind the number of equations you have and variables you are using to solve the problem. Once you have the same number of each, take a deep breath, make some substitutions, and let algebra work its magic. Finally, don’t forget to answer the question that the problem asks! You would be astounded by how often students solve a problem perfectly and then give the length when the problem asked for the width.

Thank you for reading this and I hope it was helpful! Feel free to leave comments about this post or tell me any other post you would want to see here. And if this was helpful, please subscribe and share this Substack!

All my best,

Jack