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# Getting ready for right triangles and trigonometry

Practicing finding right triangle side lengths with the Pythagorean theorem, rewriting square root expressions, and visualizing right triangles in context helps us get ready to learn about right triangles and trigonometry.
Let’s refresh some concepts that will come in handy as you start the right triangles and trigonometry unit of the high school geometry course. You’ll see a summary of each concept, along with a sample item, links for more practice, and some info about why you will need the concept for the unit ahead.
This article only includes concepts from earlier courses. There are also concepts within this high school geometry course that are important to understanding right triangles and trigonometry. If you have not yet mastered the Introduction to triangle similarity lesson, it may be helpful for you to review that before going farther into the unit ahead.

## Pythagorean theorem

### What is this, and why do we need it?

The Pythagorean theorem is ${a}^{2}+{b}^{2}={c}^{2}$, where $a$ and $b$ are lengths of the legs of a right triangle and $c$ is the length of the hypotenuse. The theorem means that if we know the lengths of any two sides of a right triangle, we can find out the length of the last side. We can find right triangles all over the place—inside of prisms and pyramids, on maps when we're finding distance, even hiding inside of equilateral triangles!

### Practice

Problem 1.1
Find the value of $x$ in the triangle shown below.

### Where will we use this?

Here are a few of the exercises where reviewing the Pythagorean theorem might be helpful:

## Simplify square root expressions

### What is this, and why do we need it?

For geometry, the square root function takes the area of a square as the input and give the length of a side of the square as an output. We'll use square root expressions when we use the Pythagorean theorem to find a side length. The trigonometric ratios for benchmark angles like $30\mathrm{°}$, $45\mathrm{°}$, and $60\mathrm{°}$ depend on square root expressions.

### Practice

Problem 2.1
Simplify.
Remove all perfect squares from inside the square root.
$\sqrt{\phantom{A}72}=$

### Where will we use this?

Here are a few of the exercises where reviewing square root expressions might be helpful.

## Visualizing right triangles in context

### What is this, and why do we need it?

Remember how there are right triangles hiding everywhere? To apply the Pythagorean theorem and trigonometry in context, we need to notice where the right angles are and think about what the hypotenuse and legs represent. Then we figure out where the measurements we have fit into the picture.

### Practice

Problem 3.1
The Memphis Pyramid in the US is a right, square-based pyramid with a height of . Each side of the base is .
Which diagram best relates the given information to the
$\ell$ of the pyramid?

We don't have an exercise for this, because the best way to practice is by drawing your own diagrams on paper or your surface of choice!

### Where will we use this?

Here are a few of the exercises where practicing visualizing right triangles might be helpful:
By the end of the unit, you should be able to find all of the unlabeled lengths and angle measurements in the diagrams, not just the ones we asked about. Come back at the end of the unit and see how much you've learned!

## Want to join the conversation?

• This is too crazy, why we need this?
• Without trig, GPS would not work, so if you want to continue to use this, you are unknowingly using trig. We need trig for a lot of practical applications in the real world whether you personally use it or not in the future, so we very much need this.
• Why is trigonometry so difficult? i dont understand anything about these triangles.
• If you do not understand anything, you need to go back to other videos. Do you understand how to identify a right triangle? Do you understand that there are two possibilities for the reference angle (one of two acute angles of right triangle)? Do you understand how to label the opposite, adjacent, and hypotenuse of a right triangle? Do you understand what SOH CAH TOA means as a mnumonic device?
I tend to doubt that you do not understand anything.
• First Question?
• To do problem 1.1, you have to use the Pythagorean theorem. If you will remember that says a^2 + b^2 = c^2, with a and b being the legs of a right triangle, meaning the two sides that share the right angle, and c being the hypotenuse (the longer side). We have two values, one leg with a value of 2, and the hypotenuse with a value of 7. To find the missing value, x, we can plug those values into the Pythagorean theorem to get 2^2 + x^2 = 7^2. We then solve to get 4 + x^2 = 49, x^2 = 45, and finally x = sqrt(45). If we were just doing an equation, it would be plus or minus square root of 45. But because we are dealing with sides and we cannot have a negative length, it is just positive.
• what does it mean when it says "remove all perfect squares from inside the square root"?
• Perfect squares include 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, ... which is a diagonal from top left to bottom right on a times table. the square root of a perfect square will be a whole number (though later you will work with decimal perfect squares also). You could look at √4 = 2 or √2^2 or √(2*2) where root and square cancel (since they are opposite operations) to give 2. When we want to simplify square roots, we want to "remove all perfect squares from inside the square root." If the number is a perfect square, then we no longer need the square root sign. However, if a number is a composite of a perfect square and a non-perfect square, the perfect square part can come out of the root, but the non-perfect number will stay in. Look at a few examples: √75 is not simplified, but if we prime factor this, we get √(5*5*3), two 5s come out as 1 to give 5√3, thus removing all perfect squares. √1250, we have to break 1250 down into √(50*25)=√(25*2*25), so the 25^2 is a perfect square which can be removed to give 25√2. √459 = √(9*51)=√9*3*17) which allows us to take out √9 to end up as 3√51.
• How is this course different from trigonometry under Pre -Calculus?
• its like an introduction
• What does it mean about the lateral edge and angle of elevation? It is very confusing. I am lost.
• So, if you're confused about a lateral edge, a lateral edge is the side edge of a shape. Basically, a side edge is shown almost external. But an angle of elevation is the degree of uplift. For example, imagine a you're looking up at a tree. If you looked straight, you altered your vision 0 degrees. However, if you looked up, it might change by 40, 50, 60 degrees. Basically, the idea here is that angle of elevation is the degree of change upwards.
• it's easy but how does squaring both side help us figure out the other length. like whats the why behind this?
• just trust the process, at the end of the day make sure you get an A