Geometry find the value of x triangle

Supplementary angles are angles whose measures sum to 180°. In the lesson below, we will review this idea along with taking a look at some example problems.

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In the image below, you see one of the common ways in which supplementary angles come up. The angles with measures \(a\)° and \(b\)° lie along a straight line. Since straight angles have measures of 180°, the angles are supplementary.

Example problems with supplementary angles

Let’s look at a few examples of how you would work with the concept of supplementary angles.

Example

In the figure, the angles lie along line \(m\). What is the value of \(x\)?

Solution

The two angles lie along a straight line, so they are supplementary. Therefore: \(x + 118 = 180\). Solving this equation:
\(\begin{align}x + 118 &= 180\\ x &= \boxed{62} \end{align} \)

Example

The angles \(A\) and \(B\) are supplementary. If \(m\angle A = (2x)^{\circ}\) and \(m\angle B = (2x-2)^{\circ}\), what is the value of \(x\)?

Solution

Since the angles are supplementary, their measures add to 180°. In other words: \(2x + (2x – 2) = 180\). Solving this equation gives the value of \(x\).

\(\begin{align}2x + (2x – 2) &= 180\\ 4x – 2 &= 180\\ 4x &= 182\\ x &= \boxed{45.5} \end{align} \)

The previous example could have asked for some different information. Let’s look at a similar example that asks a slightly different question.

Example

The angles \(A\) and \(B\) are supplementary. If \(m\angle A = (2x+5)^{\circ}\) and \(m\angle B = (x-20)^{\circ}\), what is \(m \angle A\)?

Solution

This time you are being asked for the measure of the angle and not just \(x\). But, the value of \(x\) is needed to find the measure of the angle. So, first set up an equation and find \(x\).

\(\begin{align}2x+5 + x – 20 &= 180\\ 3x-15 &= 180 \\ 3x &= 195\\ x&= 65\end{align}\)

The measure of angle \(A\) is then:
\(m\angle A = (2x+5)^{\circ}\) and \(x = 65\)

\(m\angle A = (2(65)+5)^{\circ} = \boxed{135^{\circ}} \)

Summary

There isn’t much to working with supplementary angles. You just have to remember that their sum is 180° and that any set of angles lying along a straight line will also be supplementary.

The circumference of a circle is the distance around the outside of a circle, which can be calculated using the radius \(r\) and the circumference formula \(C = 2\pi r\). Since it is a distance, the answer is given using any unit of length such as feet, inches, meters, or kilometers. In the lesson below, we will look at the different types of problems that use the circumference, including word problems.

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Examples of finding the circumference of a circle

First, we will consider some mathematical examples where you are given the radius or the diameter and simply asked to find the circumference. As with area questions, it is important to pay attention to which value you are given since this affects the calculation (remember the diameter is always twice the radius).

Let’s do an example of each of the two cases and see how the formula is applied in the different situations.

Example (given radius)

The radius of a circle is 3 cm. What is the circumference of this circle?

Solution

Apply the circumference formula with \(r = 3\).

\(\begin{align}C &= 2\pi r\\ &= 2\pi(3)\\ &= \bbox[border: 1px solid black; padding: 2px]{6\pi \text{ cm}}\end{align}\)

You could also use \(\pi = 3.14\) and approximate the answer as \(6 \times 3.14 = 18.84 \text{ cm}\).

Example (given diameter)

A circle has a diameter of 10 feet. What is the circle’s circumference?

Solution

Remember that the radius of a circle is half the diameter. Therefore, you will use \(r = 5\text{ feet}\) in the circumference formula.

\(\begin{align}C &= 2\pi r\\ &= 2\pi(5) \\ &= \bbox[border: 1px solid black; padding: 2px]{10\pi \text{ feet}}\end{align}\)

As before, you could approximate this as \(10 \times 3.14 = 31.4 \text{ feet}\).

These examples were the types you will see when you are first learning how to use this formula. But it is also possible to solve applied and other types of word problems using the circumference formula. We will look at those next.

Word problems using the circumference

To solve any type of applied problem in geometry, you must be able to identify the values you need to find and those you are given. This is based on your understanding of geometry ideas like area, perimeter, and volume – or really any idea – depending on the problem.

When it comes to circumference, that means understanding that it represents the distance around the circle. So, if you are given a real-life situation involving finding a distance like this you know to apply the circumference formula.

Example

A group has raised money to build a memorial fountain in the center of town. The fountain will be circular with a small statue in the center. The distance from the center of the statue to the outer edge of the fountain will be 15 feet. On the day the fountain is complete, the group will place a ribbon around the outside of the circle to be cut in a ceremony. How many feet of ribbon will they need? Assume that the ribbon is placed 6 inches outside of the edge of the fountain, all the way around.

Solution

If the statue is at the center of the circle, then the distance from the statue to the outside of the fountain is the radius. But the ribbon will enclose a circle that has a radius 6 inches, or half a foot, larger than this. Therefore we should use:

\(r = 15.5 \text{ feet}\)

Now we can apply the formula to find:

\(\begin{align}C &= 2\pi r \\&= 2\pi(15.5)\\ &= 31\pi\\ &\approx 31 \times 3.14 \\&= \bbox[border: 1px solid black; padding: 2px]{97.3 \text{ feet}}\end{align}\)

Now while this example used a real-life situation, some word problems are a more mathematical in nature and may require you to use your knowledge of algebra. This is true in the next example, where you are given the circumference, but must find the diameter of the circle.

Example

The circumference of a circle is 3 meters. In terms of \(\pi\), what is the diameter of this circle?

Solution

With the information given, we know that:

\(C = 3 \text{ meters} = 2\pi r\)

We can solve for \(r\) by dividing both sides by \(2 \pi\).

\(\dfrac{3}{2\pi} = r\)

However, the question asks us to find the diameter, \(d\). This is twice the radius. Therefore:

\(\begin{align}d &= 2r\\ &= 2 \times \dfrac{3}{2\pi}\\ &= \bbox[border: 1px solid black; padding: 2px]{\dfrac{3}{\pi} \text{ meters}}\end{align}\)

You wouldn’t round here since the question asked for the value “in terms of \(\pi\);”.

Summary

Circumference is really just another way of talking about the perimeter, or the distance around, a circle. This can be calculated in exact form (in terms of \(\pi\) or as a decimal approximation using the circumference formula. From there you can also solve applied problems or more complicated mathematical problems using your general understanding of circles and their properties.

Continue studying circles

You can review how to find the area of a circle in the following lesson: Finding the area of a circle

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

Table of Contents

1. Examples of using the Pythagorean theorem
2. Solving applied problems (word problems)
3. Solving algebraic problems
4. Summary

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Applying the Pythagorean theorem (examples)

In the examples below, we will see how to apply this rule to find any side of a right triangle triangle. As in the formula below, we will let a and b be the lengths of the legs and c be the length of the hypotenuse. Remember though, that you could use any variables to represent these lengths.

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.

Example

Find the value of \(x\).

Solution

The side opposite the right angle is the side labelled \(x\). This is the hypotenuse. When applying the Pythagorean theorem, this squared is equal to the sum of the other two sides squared. Mathematically, this means:

\(6^2 + 8^2 = x^2\)

Which is the same as:

\(100 = x^2\)

Therefore, we can write:

\(\begin{align}x &= \sqrt{100}\\ &= \bbox[border: 1px solid black; padding: 2px]{10}\end{align}\)

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.

In other situations, you will be trying to find the length of one of the legs of a right triangle. You can still use the Pythagorean theorem in these types of problems, but you will need to be careful about the order you use the values in the formula.

Example

Find the value of \(y\).

Solution

The side opposite the right angle has a length of 12. Therefore, we will write:

\(8^2 + y^2 = 12^2\)

This is the same as:

\(64 + y^2 = 144\)

Subtracting 64 from both sides:

\(y^2 = 80\)

Therefore:

\(\begin{align}y &= \sqrt{80} \\ &= \sqrt{16 \times 5} \\ &= \bbox[border: 1px solid black; padding: 2px]{4\sqrt{5}}\end{align}\)

In this last example, we left the answer in exact form instead of finding a decimal approximation. This is common unless you are working on an applied problem.

Applications (word problems) with the Pythagorean theorem

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.

Example

Two hikers leave a cabin at the same time, one heading due south and the other headed due west. After one hour, the hiker walking south has covered 2.8 miles and the hiker walking west has covered 3.1 miles. At that moment, what is the shortest distance between the two hikers?

Solution

First, sketch a picture of the information given. Label any unknown value with a variable name, like x.

Due south and due west form a right angle, and the shortest distance between any two points is a straight line. 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. You will likely need to round your answer.

\(\begin{align}x &= \sqrt{17.45} \\ &\approx 4.18 \text{ miles}\end{align}\)

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.

Algebra style problems with the Pythagorean theorem

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.

Example

A right triangle has a hypotenuse of length \(2x\), a leg of length \(x\), and a leg of length y. Write an expression that shows the value of \(y\) in terms of \(x\).

Solution

Since no figure was given, your first step should be to draw one. The order of the legs isn’t important, but remember that the hypotenuse is opposite the right angle.

Now you can apply the Pythagorean theorem to write:

\(x^2 + y^2 = (2x)^2\)

Squaring the right-hand side:

\(x^2 + y^2 = 4x^2\)

When the problem says “the value of \(y\)”, it means you must solve for \(y\). Therefore, we will write:

\(y^2 = 4x^2 – x^2\)

Combining like terms:

\(y^2 = 3x^2\)

Now, use the square root to write:

\(y = \sqrt{3x^2}\)

Finally, this simplifies to give us the expression we are looking for:

\(y = \bbox[border: 1px solid black; padding: 2px]{x\sqrt{3x}}\)

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Summary

The Pythagorean theorem allows you to find the length of any of the three sides of a right triangle. 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. the hypotenuse and to always draw a picture when one isn’t given.

The area of a circle can be thought of as the number of square units of space the circle occupies. This can be found using either the radius or the diameter, which we will cover in the examples below. We will also look at some examples of word problems involving area that you may come across in your studies.

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Examples of finding the area of a circle

We will use the following formula to find the area of any circle. Notice that this formula uses the radius, so we will have to convert when we are given the diameter instead. Let’s look at both cases.

Example (given radius)

Find the area of a circle with a radius of 5 meters.

Solution

Apply the formula: \(A = \pi r^2\) with radius \(r = 5\). Remember that \(\pi\) is about 3.14.

\(\begin{align}A &= \pi (5)^2 \\ &= 25\pi \\ &\approx \bbox[border: 1px solid black; padding: 2px]{75.5 \text{ m}^2}\end{align}\)

A few comments about this final answer.

Since the units of the radius were in meters, the answer is in square meters. This can be written out in words, or as \(\text{m}^2\). Also, the final answer can be written in terms of \(\pi\) (\(25 \pi\) square meters) or as a decimal approximation (75.5 square meters). Which one you use depends on the application and the problem you are working on.

Example (given diameter)

Find the area of a circle with a diameter of 6 feet.

Solution

The radius of any circle is always half the diameter. Since the diameter of the circle is 6 feet, the radius must be 3 feet (the radius is always half of the diameter). So, we can apply the formula using \(r = 3\).

\(\begin{align}A &= \pi (3)^2 \\ &= 9\pi \\ &\approx \bbox[border: 1px solid black; padding: 2px]{28.3 \text{ ft}^2}\end{align}\)

As you can see, it is important to pay attention to whether or not you are given the radius or the diameter of the circle. In some word problems though, this may not always be as clear.

Word problems involving the area of a circle

Not every problem you will encounter will simply say “find the area”. In the next two examples, you will see other types of questions you might be asked.

Example

Jason is painting a large circle on one wall of his new apartment. The largest distance across the circle will be 8 feet. Approximately how many square feet of wall will the circle cover?

Solution

Whenever you are asked to find the number of square feet covered by something, you are finding an area. To find the area of Jason’s circle, we first need to figure out if we have been given the radius or the diameter. By definition, the diameter of a circle is the longest distance across the circle, so we know here that the diameter is 8 feet. This means that the radius is 4 feet. Therefore:

\(\begin{align}A &= \pi (4)^2 \\ &= 16\pi \\ &\approx \bbox[border: 1px solid black; padding: 2px]{50.2 \text{ square feet}}\end{align}\)

So, Jason’s circle will cover about 50.2 square feet of his wall.

Example

The area of a circle is \(81 \pi\) square units. What is the radius of this circle?

Solution

To answer this question, you will have to remember a little bit of algebra. Use the formula and substitute the values you know. Then, solve for the radius, \(r\).

Start with the area formula.

\(A = \pi r^2\)

Substitute in \(A = 81 \pi\) since you know this is the area.

\(81\pi = \pi r^2\)

Divide both sides by \(\pi\).

\(81 = r^2\)

This can also be written as:

\(r^2 = 81\)

Take the square root to find \(r\). Since this is a radius, the value of \(r\) must be positive.

\(\begin{align}r &= \sqrt{81} \\ &= \bbox[border: 1px solid black; padding: 2px]{9}\end{align}\)

Therefore, the radius must be 9.

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Summary

Finding the area of a circle is all about applying the formula. But be careful! You always must pay attention to the information you are given – especially in applied problems.