Perimeter, Right Triangles and Radicals

In my Algebra class today, we were reviewing the process for adding and subtracting radical numbers. Previously, we simplified irrational numbers, such as the square of 80, which becomes 4 times the square root of 5.

To give them a more thoughtful question than just what is the sum of SQRT(75) + SQRT(48), where the only "thought" is to get past thinking that it's SQRT(123), I decided to pull out Right Triangles and that Old Favorite, the Pythagorean Theorem. Now, I didn't want them to just simplify the irrational number, I wanted some kind of addition in the problem. That brings us to Perimeter of a Right Triangle.

There are basically two types of problems you can offer up for consideration: problems with one radical number, and problems with two radical numbers. Three is just being mean, and overly complicates things -- on the other hand, it could make it interesting. Give the students the length of the legs (the base and the height), establish that there is a right angle between them, and let them go to work.

In the first kind of problem, you can pick any two numbers and you'll most likely have an irrational hypotenuse. Okay, but boring. It's more interesting if you make one of the legs irrational in such a way to make the hypotenuse rational. Surprise them.

It keeps it interesting when you consider these two problems, which look very similar, but are very different.

Consider the square root of 13 triangle first. If we square 6, we get 36. If we take the square of the square root of 13, we get 13. 36 + 13 = 49, which is the square of the hypotenuse. Therefore, the hypotenuse is 7.

Finding the perimeter is as simple as adding the three sides, which in this case means combining the like terms, which would be the integers 6 and 7. The perimeter is 13 + root 13.

In the square root 12 problem, the numbers had to be carefully selected. In this case, there will be two irrational numbers. If anything is to be combined, then the radicals have to simplify to the same radicand. Otherwise, the exercise is pointless.

If we square 6, we get 36 again. If we take the square of the square root of 12, we get 12, of course. 36 + 12 = 48, which is the square of the hypotenuse. Therefore, the hypotenuse is root 48.

We can't add any of the numbers as they are written, but we can simplify both of the radicals, as shown:

Both of the radicals have the square root of 3 in simplest form, and their coefficients can be added. The perimeter is 6 + 6 square root 3.

Interestingly, in both examples I wrote for lessons today, a double number appeared in the answer. This is actually something else to be careful about. Some students might see that as a co-incidence. Others might see it as a pattern and come to expect it.

There is an easy solution to that: give them some more problems to work on!