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How to Use the Pythagorean Theorem

Updated on June 6, 2015

Let's Begin

The very first thing you must remember when you attempt to use this theorem is that IT ONLY WORKS RIGHT TRIANGLES, that is triangles where one of the three angles are 90% (like the corner of a square or a rectangle. This will not yield accurate results on any other type of triangle such as an equilateral, isosceles, or scalene triangle.

Now the lengths of a right triangle maintain a consistent proportion. Consider for a moment if one person travels directly north for a distance X and stops and another person travels directly to the west for a distance Y and stops (these lines of travel represent the legs of the triangle) the straight line that connects their end points (this is the hypotenuse or length C) must have a constant relation to the lengths of the legs (lengths A and B). If this does not quite intuitively make sense don't fret it doesn't need to in order to solve these problems. Just read on.


The Albegra

So the Pythagorean theorem says that A2 + B2 = C2. What does this mean? It Quite simply means that whatever the distance of leg A is squared plus the distance of leg B squared will always be the distance of the hypotenuse (the long side) C squared.

Let's looks at a simple Example to check that this is true.

Application

So for the legs of this right triangle we have lengths 3 and 4. The hypotenuse is length 5. So does 32+42=52?Three squared equals 9 plus four squared equals 16 equals 25. And 5 square equals 25 so the formula works and it always works. The numbers might not be so neat but the theorem only holds.

Now Pythagorean Theorem problems will take two forms;

In the first form you will be given the lengths of A and B and asked to find length C

Here's an Example

Right triangle X has three sides. The shorter two side measure 5 units and 12 units. What is the length of the longest side.

Solution: We are told that the shorter two sides are 5 and 12, these are the legs and are represented by A and B in the equation so we plug these values in and have;

52+122=C2. C is our unknown and the length we are solving for.

52=25

122=144 so,

25+144=C2 or 169=C2

to solve 169=C2 we need to get C by itself by getting rid of the square that it is being raised to. Remember in algebra whatever you do to one side of an equation you must also do to the other side of the equation! The inverse operation of a square is a root so we will take the root of each side.

√169=√C2 This works out to

13=C

And you are done. Though the number often won't be this neat just use the square root key on your calculator and the answer will be correct.

The next type of Pythagorean theorem question involves only slightly more algebra. You will be given the length of the hypotenuse and the length of one leg and asked to find the remaining leg.

For example; Right Triangle Y has hypotenuse of length 10 and another side of length 8. What is the third side?

Solution; Again plug what you know into the formula

A2+B2=C2

We know that C, the hypotenuse is length 10 and one of the legs is length 8, we call plug the 8 into A or B it doesn't matter. So,

82+B2=102 or

64+B2=100

Again we must get what we are solving for B by itself, so first we will deal with getting the 64 on the other side of the equation. The inverse of adding 64 is subtracting 64 so we will do this to both sides of the equation.

64-64+B2=100-64. 64 cancels on the left side leaving us with B2 and reduces the right side to 36.

B2=36 is what we are left with. We will now repeat the process of eliminating the square of B by taking the root of both sides just as we did in the first problem.

√B2=√36

This works out to B=6 and we have found our answer, the length of the unknown leg.

Hope this helps

I hope this brief tutorial clears up any problems people might be having with Pythagorean theorem problems. Remember the numbers won't usually be as tidy as in the examples I used but the formula will always work for right triangles. If you need help on any specific question leave them in the comments section and I'll try to get to them. Thanks and happy calculating!!!

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    • Larry Rankin profile image

      Larry Rankin 2 years ago from Oklahoma

      Very useful. I always loved geometry, and this theorem is one of the most useful.

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