Pythagoras Theorem Problem Solving

In this lesson, we will look at several different types of examples of applying this theorem.

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There are many different kinds of real-life problems that can be solved using the Pythagorean theorem.

The easiest way to see that you should be applying this theorem is by drawing a picture of whatever situation is described.

The position Z is situated at the north of X and at a distance of 60 m from X. PS = 80 Therefore perimeter of the rectangle PQRS = 2 (length width) = 2 (150 80) m = 2 (230) m = 460 m 4.

A ladder 13 m long is placed on the ground in such a way that it touches the top of a vertical wall 12 m high.

There is one last type of problem you might run into where you use the Pythagorean theorem to write some type of algebraic expression.

This is something that you will not need to do in every course, but it does come up.Therefore, we can apply the Pythagorean theorem and write: \(3.1^2 2.8^2 = x^2\) Here, you will need to use a calculator to simplify the left-hand side: \(17.45 = x^2\) Now use your calculator to take the square root. \(\beginx &= \sqrt \ &\approx 4.18 \text\end\) As you can see, it will be up to you to determine that a right angle is part of the situation given in the word problem.If it isn’t, then you can’t use the Pythagorean theorem.In each example, pay close attention to the information given and what we are trying to find.This helps you determine the correct values to use in the different parts of the formula. The side opposite the right angle is the side labelled \(x\). When applying the Pythagorean theorem, this squared is equal to the sum of the other two sides squared.The Pythagorean theorem is a way of relating the leg lengths of a right triangle to the length of the hypotenuse, which is the side opposite the right angle.Even though it is written in these terms, it can be used to find any of the side as long as you know the lengths of the other two sides.Mathematically, this means: \(6^2 8^2 = x^2\) Which is the same as: \(100 = x^2\) Therefore, we can write: \(\beginx &= \sqrt\ &= \bbox[border: 1px solid black; padding: 2px]\end\) Maybe you remember that in an equation like this, \(x\) could also be –10, since –10 squared is also 100.But, the length of any side of a triangle can never be negative and therefore we only consider the positive square root.It is one of those things that you should memorize, as it comes up in all areas of math, and therefore in many different math courses you will probably take.Remember to avoid the common mistake of mixing up where the legs go in the formula vs.

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