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Quadratic Formula Examples—Solved Step-by-Step

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Quadratic Formula Examples—Solved Step-by-Step

Quadratic Formula Examples Tutorial

Step-by-Step Guide: Examples of How to Find the Roots of a Quadratic Function using the Quadratic Formula

 

Step-by-Step Guide: Quadratic Formula Examples Solved

 

Are you ready to work through a few quadratic formula examples to gain some more practice and experience with solving quadratic equations using the quadratic formula?

In math, the quadratic formula, x= (-b ± [√(b² - 4ac)]) / 2a is an incredibly important and useful formula that you can use to find the solutions (also known as roots) or any quadratic equation of the form ax² + bx + c = 0 (where a ≠ 0), whether it is easy to factor or not!

If you know how to use the quadratic formula, then you can solve a variety of algebra problems involving quadratic equations, and learning how to use it correctly is something that you can easily learn with some practice and repetition.

This free Quadratic Formula Examples Step-by-Step Guide includes a short review of the quadratic formula as well as several different practice problems that we will work through and solve using the quadratic formula with a step-by-step explanation. The guide is organize by the following sections, and you can click on any of the hyperlinks below to jump to any particular spot:

Before we dive into any of the quadratic formula examples, let’s start off with a quick review of the quadratic formula and why it is such a useful algebra tool.

 

Figure 01: The Quadratic Formula

 

What is the Quadratic Formula?

Before you can learn how to use the quadratic formula, it is important that you understand what a quadratic equation is.

Definition: A quadratic equation is a function of the form ax² + bx + c = 0 (where a does not equal zero). On a graph, a quadratic equation can be represented by a parabola. The x-values where the parabola crosses the x-axis is called the solutions, or roots, of the quadratic equation.

For example, consider the following quadratic equation:

  • x² + 5x + 6 = 0

Notice that this equation is in ax² + bx + c = 0 form, where…

  • a=1

  • b=5

  • c=6

If we want to find the solutions, or roots, of this quadratic equation, we have a few options.

First, we could factor this quadratic equation by looking for two values that add to 5 and also multiply to 6, which, in this case, would be 2 and 3. So we could say that…

  • x² + 5x + 6 = 0 → (x+2)(x+3) = 0

We could then solve for each factor as follows:

  • x + 2 = 0 → x = -2

  • x + 3 = 0 → x = -3

Now we can conclude that the solutions of this quadratic are x=-2 or x=-3.

 

Figure 02: What are the solutions (or roots) of a quadratic equation?

 

Another option for finding the solutions to a quadratic equation is to look at its graph. The solutions, or roots, will be the x-values where the graph crosses the x—axis. Note that quadratic equations can have two roots, one root, or even no real roots (as you will see later in this guide).

As for the equation x² + 5x + 6 = 0, the corresponding graph in Figure 02 above confirms that the equation has solutions at x=-2 and x=-3.

But what do we do when a quadratic equation is very difficult to factor or when we do not have access to a clear graph? Well, this is where the quadratic formula comes into play.

Definition: Any quadratic equation of the form ax² + bx + c = 0 (where a ≠ 0), can be solved using the quadratic formula, which states that…

  • x= (-b ± [√(b² - 4ac)]) / 2a

Why is the quadratic formula so useful? Because, as the definition states, it can be used to find the solutions to any quadratic equation. While the quadratic equation that we just looked at, x² + 5x + 6 = 0, was pretty easy to work with and solve, it is considered extremely simple. As you move farther along your algebra journey, you will come across more and more complex quadratic equations that can be very difficult to factor or even graph.

However, if you know how to use the quadratic formula, you can successfully solve any quadratic equation. With this in mind, let’s go ahead and work through some quadratic formula examples so you can gain some practice.

And we will start by using it to solve x² + 5x + 6 = 0, because we already know that the solutions are x=-2 and x=-3. If the quadratic formula works, then it should yield us that same result. Once we work through this first simple example, we will move onto more complex examples of how to use the quadratic formula to solve quadratic equations.

 

Figure 03: To use the quadratic formula, start by identifying the values of a, b, and c.

 

Quadratic Formula Examples

We will begin by using the quadratic formula to solve the equation shown in Figure 02 above: x² + 5x + 6 = 0

Example #1: Solve x² + 5x + 6 = 0

First, notice that our equation is in ax² + bx + c = 0 form where:

  • a=1

  • b=5

  • c=6

Identifying the values of a, b, and c will always be the first step (provided that the equation is already in ax² + bx + c = 0 form).

Now that we know the values of a, b, and c, we can plug them into the quadratic equation as follows:

  • x= (-b ± [√(b² - 4ac)]) / 2a

  • x= -(5) ± [√(5² - 4(1)(6))]) / 2(1)

  • x= -5 ± [√(25 - 24)] / 2

  • x= -5 ± [√(1)] / 2

  • x= (-5 ± 1) / 2

Now we are left with x= (-5 ± 1) / 2. Note that the ± mean “plus or minus” meaning that we have to split this result into two separate equations:

  • Plus: x = (-5 + 1) / 2

  • Minus: x = (-5 - 1) / 2

By solving these two separate equations, we can find the solutions to the quadratic function x² + 5x + 6 = 0.

  • x = (-5 + 1) / 2 = -4/2 = -2 x=-2

  • x = (-5 - 1) / 2 = -6/2 = -3 x=-3

After solving both equations, we are left with x=-2 and x=-3, which we already knew were the solutions to x² + 5x + 6 = 0. So, we have confirmed that the quadratic formula can be used to find the solutions to any quadratic equation of the form ax² + bx + c = 0.

Final Answer: x=-2 and x=-3

The steps to solving the quadratic formula example is illustrated in Figure 04 below.

 

Figure 04: Quadratic Formula Examples Step-by-Step

 

Example #2: Solve 2x² + 2x -12 = 0

For our next quadratic formula example, we will again start by identifying the values of a, b, and c as follows:

  • a=2

  • b=2

  • c=-12

Make sure that you correctly identify the sign (positive or negative) as well, since this is necessary to using to quadratic formula correctly.

Next, we can substitute these values for a, b, and c into the quadratic formula as follows:

  • x= (-b ± [√(b² - 4ac)]) / 2a

  • x= -(2) ± [√(2² - 4(2)(-12))]) / 2(2)

  • x= -2 ± [√(4 - -96)] / 4

  • x= -2 ± [√(100)] / 4

  • x= (-2 ± 10) / 4

Our result is x= (-2 ± 10) / 4. From here, we can rewrite e the result as two separate equations by “spitting” the ± sign as follows:

  • Plus: x= (-2 + 10) / 4

  • Minus:x= (-2 - 10) / 4

Now we can solve each individual equation to find the values of x that will be the solutions of this quadratic equation.

  • x= (-2 + 10) / 4 = 8/4 = 2 x=2

  • x = (-2 - 10) / 4 = -12/4 = 3 x=3

We are left with two values for x: x=2 and x=-3, and we can conclude that the quadratic equation 2x² + 2x -12 = 0 has the following solutions:

Final Answer: x=2 and x=-3

Figure 05 shows the step-by-step process for solving this quadratic formula example.

 

Figure 05: Quadratic Formula Examples #2 Solved

 

Example #3: Solve 2x² -5x + 3 = 0

For the next of our quadratic formula examples calls for us to use the quadratic formula to find the solutions to a quadratic function where:

  • a=2

  • b=-5

  • c=3

The process of substituting a, b, and c into quadratic formula will be exactly the same as the last two quadratic formula examples.

  • x= (-b ± [√(b² - 4ac)]) / 2a

  • x= -(-5) ± [√(-5² - 4(2)(3))]) / 2(2)

  • x= 5 ± [√(25 - 24)] / 4

  • x= 5 ± [√(1)] / 4

  • x= (5 ± 1) / 4

Are you starting to get the hang of it? Now that we have simplified our equation, we are left with x= (5 ± 1) / 4. And, just like the last two examples, we can go ahead and split this result into two separate equations as follows:

  • Plus: x= (5 + 1) / 4

  • Minus: x= (5 - 1) / 4

Finally, we just have to solve each equation to get our final answer (i.e. the values of the solutions).

  • x= (5+1) / 4 = 6/4 = 3/2 x=3/2

  • x= (5-1) / 4 = 4/4 = 1 x=1

Notice that the result of the first equation ended up as a fraction (3/2). This is totally fine! It just means that the parabola will cross the x-axis in the middle of a box (rather than hitting directly at an integer coordinate).

Final Answer: x=3/2 and x=1

All of the steps for solving this example are shown in Figure 06 below.

 

Figure 06: Sometimes a quadratic formula will give you a solution that is a fraction.

 

Example #4: Solve 3x² + 2 = 7x

The fourth and final of our quadratic formula examples looks a bit different. The given equation 3x² + 2 = 7x is not in ax² + bx + c = 0 form.

Whenever this is the case, we will have to see if we can use algebra to rearrange the equation so to make into ax² + bx + c = 0 form. We can do that by using inverse operations to move the 7x to the left-side of the equation as follows:

  • 3x² + 2 = 7x

  • 3x² + 2 (-7x) = 7x (-7x)

  • 3x² + 2 -7x = 0

Notice that result, 3x² + 2 -7x = 0, still isn’t in ax² + bx + c = 0 form. However, the commutative property allows us to rearrange the terms as follows:

  • 3x² + 2 -7x = → 3x² -7x +2

Now we have an equivalent equation, 3x² -7x +2=0, that is in ax² + bx + c = 0 form, where:

  • a=3

  • b=-7

  • c=2

Sometimes you will be given equations that have to be rearranged in order to use the quadratic formula. If you can not rearrange an equation so that it can be expressed in ax² + bx + c = 0 form, then you can not solve it using the quadratic formula.

This example, however, can now be solved using the quadratic formula as follows:

  • x= (-b ± [√(b² - 4ac)]) / 2a

  • x= -(-7) ± [√(-7² - 4(3)(2))]) / 2(3)

  • x= 7 ± [√(49 - 24)] / 6

  • x= 7 ± [√(25)] / 6

  • x= (7 ± 5) / 6

Now we are left with a much easier equation to work with: x= (7 ± 5) / 6. Let’s go ahead and split it into two separate equations to solve it:

  • Plus: x= (7 + 5) / 6

  • Minus: x= (7 - 5) / 6

We can solve for x in each equation as follows:

  • x= (7+5) / 6 = 12/6 = 2 x=2

  • x= (7-5) / 6 = 2/6 = 1/3 x=1/3

Final Answer: x=2 and x=1/3

That’s all that there is to it! You can review of the steps to solving this quadratic formula example by looking at the illustration in Figure 07 below.

 

Figure 07: Quadratic Formula Examples: Rearranging an equation to put it into ax² + bx + c = 0 form.

 

Do you need more practice with using the Quadratic Formula?

Check out our free library of Quadratic Formula Worksheets (with answer keys)


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How to Complete the Square in 3 Easy Steps

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How to Complete the Square in 3 Easy Steps

How to Complete the Square in 3 Easy Steps

Step-by-Step Guide: How to do Completing the Square

 

Step-by-Step Guide: How to Complete the Square Explained.

 

As you continue onto more advanced problems where you have to factor quadratics, you will have to learn how to complete the square in order to find correct solutions. Completing the square is a special technique that you can use to factor quadratic functions.

This free step-by-step guide on How to Complete the Square will teach you an easy 3-step method for factoring any quadratic using a technique called “completing the square.”

This guide will focus on the following topics and sections. You can click on any of the text links below to jump to one particular section, or you can follow each section in sequential order.

Let’s begin by exploring the meaning of completing the square and when you can use it to help you to factor a quadratic function.

 

Figure 01: Are you ready to learn how to find solutions to quadratic equations by completing the square?

 

What is Completing the Square?

Completing the square is a method that you can use to solve quadratic equations of the form ax² + bx + c = 0 (where a, b, and c are all not equal to zero).

❗Note that the equations of the form ax² + bx + c = 0 are called quadratic equations and they can be rewritten as follows:

  • ax² + bx + c = 0

  • ax² + bx = -c

Both of these equations are equivalent to each other, and understanding the relationship between these two equations will help you to understand how to complete the square later on in this guide.

In the next section, we will work through three examples of how to complete the square using the following 3-step method:

  • Step #1: Rearrange the quadratic equation so that all of the constants are on one side of the equals sign.

  • Step #2: Add (b/2)² to both sides of the equal sign.

  • Step #3: Factor and solve.

By solving a quadratic equation by completing the square, you are identifying values where the parabola that represents the equation crosses the x-axis.

As long as you understand how to follow and apply these three steps, you will be able to solve quadratics by completing the square (provided that they are solvable). Now, let’s gain some experience with using the three step method on how to complete the square by working through some step-by-step practice problems.

 

Figure 02: The solutions to a quadratic equation are the values where the graph crosses the x-axis.

 

How to Complete the Square: Example #1

Solve: x² - 6x -16 = 0

For this first example (and all of the practice problems in this guide), we can solve the problem by completing the square using our three step method as follows:

Step #1: Rearrange the quadratic equation so that all of the constants are on one side of the equals sign.

Let’s start off by noticing that our given quadratic function is indeed in ax² +bx + c = 0 form, where a=1, b=-6, and c=-16.

To complete the first step, we have to move all of the constants (all of the values not attached to variables, to the right side of the equals sign as follows:

  • x² - 6x -16 = 0

  • x² - 6x -16 (+16) = 0 (+16)

  • x² - 6x = 16

Now we have completed the first step and we are left with a new equivalent equation:

  • x² - 6x = 16

Step #2: Add (b/2)² to both sides of the equal sign.

For the next step, we have to find the value of (b/2)² and add it to both sides of the equals sign.

Since we know that b=-6, we can find the value of (b/2)² by substituting -6 for b as follows:

  • (b/2)²

  • (-6/2)²

  • (-3)²

  • =9

In this case, (b/2)² = 9, so, to complete Step #2, we simply have to add 9 to both sides of the equal sign as follows:

  • x² - 6x = 16

  • x² - 6x + 9 = 16 + 9

  • x² - 6x + 9 = 25

 

Figure 03: How to Complete the Square in 3 Easy Steps.

 

Step #3: Factor and solve.

Finally, we are ready for the third and final step where we just need to factor and solve.

Notice that the left side of the equation of x² - 6x + 9 = 25 is a trinomial that is factorable as follows:

  • x² - 6x + 9 = (x-3)(x-3)

  • x² - 6x + 9 = (x-3)²

In this example, the factors of x² - 6x + 9 are (x-3)(x-3), which we will express as (x-3)² since it will allow us to solve the problem as follows:

  • x² - 6x + 9 = 25

  • (x-3)² = 25

  • √[(x-3)²] = √[25]

  • x -3 = ± 5

In the third step above, we took the square root of both sides of the equation to remove the exponent and we are left with x -3 = ± 5, which means that

  • x - 3 = 5

  • x - 3 = -5

We can now find the solutions to this first example by solving both equations as follows:

  • x - 3 = 5 → x = 8

  • x - 3 = -5 → x = -2

All three steps for how to do completing the square are shown in Figure 03 above.

Now, we can conclude that the original quadratic equation x² - 6x -16 = 0 has two solutions:

Final Answer: x = 8 and x = -2

This means that the graph of the equation x² - 6x -16 = 0 will be a parabola that crosses the x-axis at both (-2,0) and (8,0) as shown in Figure 04 below.

 

Figure 04: How to solve by completing the square: graph explanation

 

How to Complete the Square: Example #2

Solve: x² +12x +32 = 0

We can solve this next example using our 3-step method, just as we did in the previous example, as follows:

Step #1: Rearrange the quadratic equation so that all of the constants are on one side of the equals sign.

Notice for our given quadratic equation, x² +12x +32 = 0, that a=1, b=12, and c=32.

Since our constant c is on the left side of the equation, we simply have to move it to the right side using inverse operations to complete Step #1.

  • x² +12x + 32 = 0

  • x² +12x + 32 (-32) = 0 (-32)

  • x² + 12x = -32

After completing the first step, we now have:

  • x² + 12x = -32

Step #2: Add (b/2)² to both sides of the equal sign.

Next, we have to add (b/2)² to both sides of our new equation.

In this example, b=12, so we can find the value of (b/2)² as follows:

  • (b/2)²

  • (12/2)²

  • (6)²

  • =36

Since (b/2)² = 36, we can complete Step #2 by adding 36 to both sides of the equation as follows:

  • x² + 12x = -32

  • x² + 12x +36 = -32 +36

  • x² + 12x +36 = 4

 

Figure 05: How to complete the square to find the solutions to a quadratic equation.

 

Step #3: Factor and solve.

For the final step, we just have to factor and solve for any potential values of x.

Just like example #1, we can finish completing the square by factoring the trinomial on the left side of the equation and then solving.

In this case, the trinomial on the left side of the equation can be factored as follows:

  • x² + 12x +36 = (x+6)(x+6)

  • x² + 12x +36 = (x+6)²

❗Note that whenever you solve a problem using the complete the square method, you will always end up with two identical factors when you complete Step #3.

Now that we know that the factors of x² + 12x +36 are equal to (x+6)², we can solve for x as follows:

  • x² + 12x +36 = 4

  • (x+6)² = 4

  • √[(x+6)²] = √[4]

  • x + 6 = ± 2

In this case, the original quadratic function x² +12x +32 = 0 will have two solutions:

  • x + 6 = 2

  • x + 6 = -2

We can determine these two solutions by solving each equation as follows:

  • x + 6 = 2 → x = -4

  • x + 6 = -2 → x = -8

The entire 3-step method for completing the square for Example #2 is shown in Figure 05 above.

Final Answer: x = -4 and x = -8

Figure 06 below shows the graph of the parabola represented by x² +12x +32, with x-intercepts at -4 and -8.

 

Figure 06: Solving a quadratic by completing the square helps you to find the x-intercepts of the parabola that represents the equation.

 

How to Complete the Square: Example #3

Solve: x² +2x -7 = 0

Are you starting to get the hang of how to complete the square? Let’s gain some more experience with this next example.

Step #1: Rearrange the quadratic equation so that all of the constants are on one side of the equals sign.

For this problem, we know that a=1, b=2, and c=-7.

For our first step, let’s rearrange the equation so that all of the constants are on the right side:

  • x² +2x -7 = 0

  • x² +2x -7 (+7) = 0 (+7)

  • x² + 2x = 7

Now we have a new equivalent function:

  • x² + 2x = 7

Step #2: Add (b/2)² to both sides of the equal sign.

Now we need to add (b/2)² to both sides of the equation. Since b=2 in this example, (b/2)² is equal to:

  • (b/2)²

  • (2/2)²

  • (1)²

  • =1

Since (b/2)² = 1, we can complete the second step by adding 1 to each side of the equation as follows:

  • x² + 2x = 7

  • x² + 2x +1 = 7 +1

  • x² + 2x +1 = 8

 

Figure 07: How to Complete the Square: Solutions will not always be rational numbers.

 

Step #3: Factor and solve.

Now we are ready to factor and solve the equation.

We have a trinomial on the left side of the equation that can be factored as follows:

  • x² + 2x +1 = (x+1)(x+1)

  • x² + 2x +1 = (x+1)²

With these factors in mind, we can solve for x as follows:

  • x² + 2x +1 = 8

  • (x+1)² = 8

  • √[(x+1)²] = √[8]

  • x +1 = ± √[8]

For this third example, the quadratic function x² +2x -7 = 0 will have two solutions:

  • x + 1 = √[8]

  • x + 1 = - √[8]

If we continue onto solving these two equations, we will see that, unlike Examples #1 and #2, we do not end up with a rational answer:

  • x + 1 = √[8] → x = -1 + √8

  • x + 1 = - √[8] → x = -1 - √8

In cases like this, you can often conclude that:

Final Answer: x = -1 + √8 and x = -1 - √8

The solutions above are considered exact answers. However, if you are trying to estimate where the parabola will cross the x-axis on the coordinate plane, you could take the problem a step further by approximating for √8 as follows:

  • √8 ≈ 2.83

  • x = -1 + √8 → x = -1 + 2.83 → x =1.83

  • x = -1 + √8 → x = -1 - 2.83 → x =-3.83

Now we have two approximate solutions for x:

  • x =1.83 and x=-3.83

What does this mean? Just like we saw in Examples #1 and #2, the solutions tell you where the graph of the parabola crosses the x-axis. In this example, the graph crosses the x-axis at approximately 1.83 and -3.83, as shown in Figure 08 below.

 

Figure 08:  How to complete the square to determine the solutions to a quadratic equation.

 

Conclusion: How to Complete the Square

When learning how to solve quadratic equations of the form ax² + bx +c=0, understanding how to complete the square to find the values where x=0 is an important and useful algebra skill that you can use to solve a variety of problems.

Whenever you have an equation in ax² + bx +c=0 form, you can solve it by following these 3-simple steps to completing the square:

  • Step #1: Rearrange the quadratic equation so that all of the constants are on one side of the equals sign.

  • Step #2: Add (b/2)² to both sides of the equal sign.

  • Step #3: Factor and solve.

Note that the above 3-step method for completing the square can be used to find the solutions of any quadratic equation of the form ax² + bx +c=0. These solutions represent the x-values where the parabola that represents the equation crosses the x-axis.

Keep Learning:

How to Find the Vertex of a Parabola in 3 Easy Steps

Learn how to find the coordinates of the vertex point of any parabola with this free step-by-step guide.


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How to Find Horizontal Asymptotes in 3 Easy Steps

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How to Find Horizontal Asymptotes in 3 Easy Steps

How to Find Horizontal Asymptotes in 3 Easy Steps

Step-by-Step Guide: How to Find the Horizontal Asymptote of a Function

 

Step-by-Step Guide: How to Find the Horizontal Asymptotes of Functions.

 

During your study of algebra, you will eventually learn how to analyze and understand the behavior functions when they are represented graphically on the coordinate plane. The graph of a function gives you a visual portrayal of how the function behaves—including the location of any potential horizontal asymptotes.

This free step-by-step guide on How to Find Horizontal Asymptotes of a Function will demonstrate and explain everything you need to know about horizontal asymptotes and how to find the horizontal asymptote of any function (assuming that the function does indeed have a horizontal asymptote).

The following sections will be covered in this guide. While we recommend that you work through each section in order, you can use the quick-links below to jump to a particular section":

Let’s start off by learning more about horizontal asymptotes and what they look like.

 

Figure 01: The horizontal asymptote of a function tells you the behavior of a function as it approaches the edges of a graph.

 

What is a Horizontal Asymptote?

Before you learn how to find horizontal asymptotes, it is important for you to understand some key foundational concepts.

In algebra, a function is an equation of the form y=f(x) that represents the relationship between two sets. A relation can be defined as a function when there is only one output value for each input value (i.e. the x-values do not repeat). Functions can be visually represented on a graph (i.e. a coordinate plane).

A horizontal asymptote of a function is a horizontal line () that tells you the behavior of a function as it approaches the edges of a graph. We can also say that horizontal asymptotes allow us to identify the “end behavior” of a function.

It is also important to note that horizontal asymptotes occur when functions are rational expressions, meaning that the function is a quotient of two polynomials (i.e. the function is a fraction where both the numerator and the denominator are polynomials.

For example, consider the following function:

  • y = (3x²)/x³

 

Figure 02: The graph of the function y = (3x²)/x³

 

Our function, y = (3x²)/x³, is the quotient of two polynomials (i.e. it is a rational expression) and its corresponding graph is shown in Figure 02 above.

Since our function is a fraction, we know that we can not have zero in the denominator (otherwise the function would be undefined at this point). So, the graph of the function, as you can see in Figure 02, does something interesting at zero.

In this example, we have a horizontal asymptote at y=0.

Notice how the graph of the function y = (3x²)/x³ gets closer and closer to the line y=0, without ever touching it, as it approaches the ends of the graph (horizontally), as shown in Figure 03 below.

 

Figure 03: The function has a horizontal asymptote at y=0.

 

If we take a look at a few points on the graph of y = (3x²)/x³ and the corresponding table, as shown in Figure 04 below, we can see that as the graph moves to the left of the coordinate plane, the y-values get closer and closer to zero. And, we can also see that as the graph moves to the right of the coordinate plane, the y-values get closer and closer to zero.

This type of observation is what we call determining the “end behavior” and it helps us to understand why the graph has a horizontal asymptote at y=0.

 

Figure 04: How to Find a Horizontal Asymptote of a Function

 

Now that you understand what a horizontal asymptote is and what it looks like, you are ready to work through a few step-by-step examples of how to find a horizontal asymptote.

Before moving forward, here are a few quick key points you should be familiar with:

  • Not all functions that are rational expressions have a horizontal asymptote.

  • It is possible for a function to have 0, 1, or 2 asymptotes (i.e. a function can have a maximum of 2 asymptotes).

  • When horizontal asymptotes are shown on a graph, they are typically drawn using a dashed line, which is what you will see in this guide.

  • The horizontal asymptote of a function is not a part of the function, and it is not a requirement to include the horizontal asymptote of a function when you graph it on the coordinate plane.

  • A horizontal asymptote can be thought of as an imaginary dashed line on the coordinate plane that helps you to visual a “gap” in a graph.

 

Figure 05: A horizontal asymptote is an imaginary line that is not a part of the graph of a function.

 

How to Find Horizontal Asymptotes Example #1

Now you are ready to learn how to find a horizontal asymptote using the following three steps:

  • Step One: Determine lim x→∞ f(x). In other words, find the limit for the function as x approaches positive ∞.

  • Step Two: Determine lim x→-∞ f(x). In other words, find the limit for the function as x approaches negative ∞.

  • Step Three: If one of or both of the limits determined above is a real number, k, then the equation of the horizontal asymptote is y=k for each unique value of k (a function can have 0, 1, or 2 horizontal asymptotes).

Now, let’s go ahead and apply these three steps to our first practice problem.

Example #1: Find the horizontal asymptote of the function f(x)=4x/(x+4)

For this first example, we can start by taking a look at the graph in Figure 05 above. By looking at the graph, we can see that the function f(x)=4x/(x+4) has one horizontal asymptote at y=4.

Let’s now apply our three steps to see if confirm the at the function has a horizontal asymptote at y=4.

Step One: Determine lim x→∞ f(x). In other words, find the limit for the function as x approaches positive ∞.

  • lim x→∞ f(x) = lim x→∞ 4x/(x+4)

  • = 4 lim x→∞ [1/(1+(4/x))]

  • = 4 (lim x→∞ (1)) / (lim x→∞ (1+(4/x))

  • = 4 ( (1)/(1))

  • =4

After completing the first step, we can see that, as the function has a horizontal asymptote at y=4 as the function approaches positive ∞ on the graph.

Next, let’s confirm that the function has the same behavior as it approaches negative ∞.

Step Two: Determine lim x→-∞ f(x). In other words, find the limit for the function as x approaches negative ∞.

  • lim x→-∞ f(x) = lim x→∞ 4x/(x+4)

  • = 4 lim x→-∞ [1/(1+(4/x))]

  • = 4 (lim x→-∞ (1)) / (lim x→∞ (1+(4/x))

  • = 4 ( (1)/(1))

  • =4

Completing the second step gives us the same result as the first step (i.e. there is a horizontal asymptote at y=4).

Step Three: If one of or both of the limits determined above is a real number, k, then the equation of the horizontal asymptote is y=k for each unique value of k (a function can have 0, 1, or 2 horizontal asymptotes).

Finally, based on our results from steps one and two, we can conclude that the function has one horizontal asymptote at y=4, as shown on the graph in Figure 06 below.

Final Answer: The function has one horizontal asymptote at y=4.

 

Figure 06: How to find a horizontal asymptote using limits.

 

How to Find Horizontal Asymptotes Example #2

Find the horizontal asymptote of the function f(x)=3ˣ+5

For this next example, we want to see if the exponential function f(x)=3ˣ+5 has any horizontal asymptotes.

We can solve this problem the same as we did the first example by using our three steps as follows:

Step One: Determine lim x→∞ f(x). In other words, find the limit for the function as x approaches positive ∞.

  • lim x→∞ f(x) = lim x→∞ 3ˣ+5

  • =lim x→∞ (3ˣ) + lim x→∞ (5)

  • lim x→∞ (3ˣ) = ∞

  • lim x→∞ (5) = 5

  • = ∞ + 5 = ∞

Notice that our result for the first step is that, as the function approaches positive ∞, the limit is ∞. And, since ∞ is not a real number, we can not yet determine whether or not this function will have any horizontal asymptotes. But we can not be completely sure until we complete the second step.

Step Two: Determine lim x→-∞ f(x. In other words, find the limit for the function as x approaches negative ∞.

  • lim x→-∞ f(x) = lim x→-∞ 3ˣ+5

  • =lim x→-∞ (3ˣ) + lim x→-∞ (5)

  • lim x→-∞ (3ˣ) = 0

  • lim x→-∞ (5) = 5

  • = 0 + 5 = 5

Our result from step number two results in a real number, so we can conclude that the function does have a horizontal asymptote at y=5.

Step Three: If one of or both of the limits determined above is a real number, k, then the equation of the horizontal asymptote is y=k for each unique value of k (a function can have 0, 1, or 2 horizontal asymptotes).

Again, even though Step One did not result in a real number limit, Step Two did, so we can conclude that:

Final Answer: The function has one horizontal asymptote at y=5.

We can confirm the location of a horizontal asymptote at y=5 for the exponential function f(x)=3ˣ+5 by looking at the completed graph in Figure 07 below.

 

Figure 07: How to find horizontal asymptotes given a function explained.

 

How to Find Horizontal Asymptotes Example #3

Find the horizontal asymptote of the function f(x)=(3x^2+x)/(x+2)

For this third and final example, we have to see if the rational function f(x)=(3x^2+x)/(x+2) has one, two, or zero horizontal asymptotes.

We can determine whether or not the function has any horizontal asymptotes by following our three steps as follows:

Step One: Determine lim x→∞ f(x). In other words, find the limit for the function as x approaches positive ∞.

  • lim x→∞ f(x) = lim x→∞ (3x^2+x)/(x+2)

  • =lim x→∞ (3x+1)/(1+(2/x))

  • = (lim x→∞ (3x+1)) / (lim x→∞(1+(2/x)))

  • lim x→∞ (3x+1) = ∞

  • lim x→∞(1+(2/x)) = 1

  • = ∞ / 1 = ∞

Since our result is not a real number, Step One does not help us to determine the possible existence of a horizontal asymptote.

Step Two: Determine lim x→-∞ f(x. In other words, find the limit for the function as x approaches negative ∞.

  • lim x→-∞ f(x) = lim x→-∞ (3x^2+x)/(x+2)

  • =lim x→-∞ (3x+1)/(1+(2/x))

  • = (lim x→-∞ (3x+1)) / (lim x→-∞(1+(2/x)))

  • lim x→-∞ (3x+1) = -∞

  • lim x→-∞(1+(2/x)) = 1

  • = -∞ / 1 = -∞

Again, our result is not a real number so we can not determine the location of a horizontal asymptote.

Step Three: If one of or both of the limits determined above is a real number, k, then the equation of the horizontal asymptote is y=k for each unique value of k (a function can have 0, 1, or 2 horizontal asymptotes).

Since neither Step One nor Step Two resulted in a real number value for k, we can conclude that the function does not have any horizontal asymptotes.

Final Answer: The function does not have any horizontal asymptotes.

The graph of the function f(x)=(3x^2+x)/(x+2) is shown in Figure 08 below. Do you notice how there is no horizontal asymptote?

 

Figure 08: How to Find Horizontal Asymptotes: Not all functions will have a horizontal asymptote.

 


Conclusion: How to Find Horizontal Asymptotes

A horizontal asymptote of a function is an imaginary horizontal line () that helps you to identify the “end behavior” of the function as it approaches the edges of a graph.

Not every function has a horizontal asymptote. Functions can have 0, 1, or 2 horizontal asymptotes.

If a function does have any horizontal asymptotes, they will be displayed as a dashed line. A horizontal asymptote is an imaginary line that is not a part of the function, and it is not a requirement to include the horizontal asymptote of a function when you graph it on the coordinate plane.

You can determine whether or not any function has horizontal asymptotes by following these three simple steps:

  • Step One: Determine lim x→∞ f(x). In other words, find the limit for the function as x approaches positive ∞.

  • Step Two: Determine lim x→-∞ f(x). In other words, find the limit for the function as x approaches negative ∞.

  • Step Three: If one of or both of the limits determined above is a real number, k, then the equation of the horizontal asymptote is y=k for each unique value of k (a function can have 0, 1, or 2 horizontal asymptotes).

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How to Find the Vertex of a Parabola in 3 Easy Steps

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How to Find the Circumference of a Circle in 3 Easy Steps

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How to Find the Circumference of a Circle in 3 Easy Steps

How to Find the Circumference of a Circle

Step-by-Step Guide: What is the circumference of a circle? How to find the circumference of a circle using the formula for circumference of a circle?

 

Free Step-by-Step Guide: How to find the circumference of a circle explained.

 

Understanding how to find the circumference of a circle is an important math skill that every student must learn.

In order to learn how to find the circumference of a circle, you need to be familiar with the concept of circumference and how to use the formula for circumference of a circle to solve problems where you are tasked with finding the length of a circle’s circumference.

This free Step-by-Step Guide on How to Find the Circumference of a Circle will teach you how to use the circle formula, C=πd, to find circumference of a circle. The guide is organized by the following sections/subtopics:

While we recommend that you work through in section in order, you can use the links above to jump to any section of interest.

Are you ready to get started? Let’s begin with a review of some important properties of circles that you will need to be familiar with in order to learn how to find the circumference of a circle.

 

Figure 01: Do you know the key properties of circles?

 

What is the Circumference of a Circle?

Before we get into what the circumference of a circle actually means, there are some key circle-related vocabulary terms that you should be familiar with:

  • A circle’s diameter is any line that passes through the center point of the circle and touches two points on the edge of the circle.

  • A circle’s radius is any line segment that runs from the center point and touches the edge of the circle.

  • A radius of a circle, r, will always be equal to half the length of its diameter, d. In other words, r=2d.

  • Conversely, the diameter of a circle, d, will always be equal to twice the length of its radius, r. In other words, d=2r.

These key properties of circles are illustrated in Figure 01 above for your reference. Make sure that you understand them well before moving on.

What is the circumference of a circle?

The circumference of a circle is the length of its outer boundary. The circumference of a circle is sometimes referred to as a perimeter of a circle and it is always measured in units such as inches, centimeters, yards, etc.

In real-world terms, you could find the circumference of a circle by taking the outer boundary of a circle and flattening it out into a straight line and then measuring the length of that straight line. Whatever the length is would be the circumference of the circle.

For example, if you have a section of rope that was 4 feet long and you curled it into a perfect circle, the circle will also have a circumference of 4 feet, as shown in Figure 02 below.

 

Figure 02: What is the circumference of a circle?

 

The biggest takeaway here is that the circumference of a circle is the total length of the outer region of the circle. You could imagine taking a circle, cutting it somewhere, and then bending it so it forms a straight line. You could then find the circumference by measuring that straight line.

The diagram in Figure 03 below further illustrates the concept of circumference in mathematical terms.

Now that you understand what the circumference of a circle represents, let’s move onto the next section, where will we learn the circumference of a circle formula.

 

Figure 03: What is the circumference of a circle?

 

Circumference of a Circle Formula: C=πd

You can find the circumference of any circle simply by using the circumference of a circle formula: C=πd (where d is the length of the diameter of the circle).

In other words, circumference equals pi times diameter.

Whenever you are using the circumference of a circle formula, you have the option of using pi (π) or an approximation of pi (π≈3.14). If you have a calculator that has a π button, we highly recommend using it instead of approximating pi as 3.14, since it will give you more accurate answers. However, both options are acceptable in most cases.

Before continuing onto the next section, be sure to remember that the d in the circumference formula represents the length of the circle’s diameter. You can use the diagram in Figure 04 below as a reference for remembering that the circumference of a circle formula is C=πd.

 

Figure 04: The circumference of a circle formula is C=πd (circumference equals pi times diameter).

 

For example, let’s say that we wanted to find the circumference of a circle with a diameter of 10 cm. Knowing that d=10 in this example, we could solve this problem by using the circumference of a circle formula as follows:

  • C=πd and d=10

  • C=π(10) or C=10π

  • C=31.41592654…

  • C=31.4 cm

So, we can conclude that the circle has an approximate circumference of 31.4 centimeters. We figured this out by substituting d with 10 in the formula to get C=π(10) or C=10π (this both mean the same thing—the product of 10 and π) and using our calculator to get C=31.41592654…, which we rounded to the nearest tenth decimal place to get C=31.4 cm.

Note that we will be expressing circumference to the nearest tenth of decimal place for all of the examples in this guide.

The step-by-step process for finding the circumference of a circle with a diameter of 10cm is shown in Figure 05 below.

 

Figure 05: How to use circumference of a circle formula.

 

Now that you know how to use the circumference formula, you are ready to learn how to find the circumference of a circle by working through a several examples.


How to Find Circumference of a Circle with Diameter

In the previous section, we learned how to find the circumference of a circle with diameter of 10cm. The step-by-step process for solving thing problem is illustrated in Figure 05 above.

As long as you know that the circle circumference formula is C=πd, you can solve any problem where you have to find the circumference of a circle with diameter given. To do this, simply substitute d with the numerical value of the length of the diameter and then multiply it by π to determine the length of the circle’s circumference.

This method is exactly how we will solve our first practice problem. Let’s go ahead and get started.

Example #1: Find the Circumference of a Circle with Diameter 17 mm

For this first example, we have to find the circumference of a circle with diameter of 17 mm.

We can solve this problem by using the circumference of a circle formula, C=πd, by inputting d=17 into the formula and solving as follows:

  • C=πd

  • C=π(17) pr C=17π

  • C=53.40707511…

  • C=54.4 mm

Final Answer: The circle has a circumference of 54.4 mm.

Finished! As long as you know the length of a circle’s diameter, you can input it into the circle circumference formula to calculate its circumference.

The step-by-step process for solving Example #1 are shown in Figure 06 below. Be sure that you are comfortable with this process before moving onto the next example.

 

Figure 06: How to Find the Circumference of a Circle Step-by-Step.

 

Example #2: Find the Circumference of a Circle with Diameter 20.6 yd

For this next example, we want to calculate the circumference of a circle with diameter 20.6 yd.

Notice that the value of the diameter, d, in this example is a decimal. However, you can still use the formula, C=πd, to solve this problem and find the circumference of the circle as follows:

  • C=πd

  • C=π(20.6) or C=20.6π

  • C=64.71680866…

  • C=64.7 yd

Final Answer: The circle has a circumference of 64.7 yd.

This process for using the circumference of a circle formula to find the circumference of a circle with a diameter of 20.6 yd is illustrated in Figure 06 below.

 

Figure 06: How to find the circumference of a circle when the diameter is given.

 

Now, let’s work through one more example of how to find circumference of a circle with diameter in a real-world context.

Example #3: Find the Circumference of a Circle with Diameter 60 meters

For this third example, we want to find the circumference of a paved sidewalk that surrounds a circular pond with a diameter that is 60 meters long.

Even though this is a real-world scenario, we can still use the circumference formula to find the answer as follows:

  • C=πd

  • C=π(60) or C=60π

  • C=188.4955592…

  • C=188.5 m

Final Answer: The paved sidewalk has a circumference of 188.5 meters.

The diagram in Figure 07 illustrates this real-world scenario and how we used our formula to figure out the circumference of the path that surrounds the circular pond.

 

Figure 07: How to find the circumference of a circle in a real-world scenario.

 

How to Find Circumference of a Circle with Radius

You probably have learned that finding the circumference of a circle when you are given the length of its diameter is a pretty straightforward process, but what happens when only know the length of the circle’s radius and not the length of its diameter?

As long as you understand the relationship between a circle’s radius and its diameter, you can easily use a given radius to figure out the length of the diameter. And this skill is extremely useful because it will allow you to use the circumference of a circle formula, C=πd, to solve circle problems where only the radius is given.

Recall from our vocabulary review at the start of this guide that a radius of a circle is any line segment that starts at the center and ends on the edge of the circle and that the radius of a circle is equal to half the length of the circle’s diameter.

  • r = d/2

Conversely, we can say that the diameter of a circle is equal to twice the length of its radius:

  • d=2r

So, if you know the length of the radius of a circle, you can use it to find the length of the diameter, d, by multiplying it by 2.

For example, if you were given a circle with a radius of 5 feet (r=5). You could take the radius and double it to find the value of the diameter, which, in this case, would be 10 ft.

  • r=5

  • d=2r

  • d=2(5)

  • d=10

The diagram in Figure 08 below illustrates this relationship between the radius and the diameter of a circle.

The key takeaway is that the diameter of a circle is equal to twice the length of its radius.

Now, let’s go ahead and apply our understanding of this relationship to another practice problem.

 

Figure 08: The diameter of a circle is equal to twice the length of its radius (d=2r).

 

Example #4: Find the Circumference of a Circle with Radius of 7 inches

In this example, we have to find the circumference of a circle with a radius of 7 inches (r=7).

Our goal is to solve this problem the same way that we solved the three previous examples where we calculated circumference using the formula C=πd.

However, the difference between Example #4 and the previous three is that, in this case, we are given the radius of the circle, r, and not the diameter.

But we can use the given radius, r=7, to find the length of the diameter, d, because we know that the diameter of a circle is equal to twice the length of its radius:

  • d=2r

  • d=2(7)

  • d=14

Therefore, if the circle has a radius of 7 inches, then it also has a diameter of 14 inches (because 7x2=14).

Now that we know the value of d (d=14 in this case), we can substitute it into the circumference of a circle formula and solve as follows:

  • C=πd

  • C=π(14) or C=14π

  • C=43.98229715

  • C=44.0 in

Final Answer: The circle has a circumference of 44 inches.

The entire step-by-step process for solving this fourth and final example are illustrated in Figure 09 below. If you are still confused about how to find the circumference of a circle, we recommend that you go back and work through the practice problems again.

 

Figure 09: How to find the circumference of a circle with a given radius.

 

How to Find the Circumference of a Circle: Conclusion

The circumference of a circle, also known as its perimeter, is the length of its outer boundary.

Understanding how to find the circumference of a circle is a foundational math skill that has many practical applications both inside and outside of the math classroom.

Whenever you have to find the circumference of a circle, you can use the formula C=πd, where d represents the lengths of the circle’s diameter, to find an answer.

If you are given the length of the circle’s diameter, you can substitute it for d in the formula and solve. However, if you only know the circle’s radius, you will have to multiply it by two (d=2r) to find the length of its diameter, d, before you can use the formula.

However, as long as you know the circumference of a circle formula and the relationship between the radius and diameter of a circle, you can easily calculate the circumference of a circle in just a few simple steps.

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How to Find the Area of a Circle in 3 Easy Steps

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How to Find the Area of a Circle in 3 Easy Steps

How to Find the Area of a Circle

Step-by-Step Guide: How to Find Area of a Circle with Radius, How to Find Area of a Circle with Diameter

 

Free Step-by-Step Guide: How to find the area of a circle explained.

 

In math, it is important to be able to calculate the area of a shape (i.e. the total space that a shape takes up) by using a formula.

When it comes to finding the area of a circle, you can easily find the area as long as you know the area of a circle formula and the length of the circle’s radius or diameter.

This free Step-by-Step Guide on How to Find the Area of a Circle will teach you how to use the circle formula, A=πr², to find area of a circle and it will cover the following topics:

You have the option of either following this guide on how to find area of a circle in step-by-step order, or you can use the hyperlinks above to skip to any particular section of interest.

Now, before we get into any examples of how to find the area of a circle, let’s do a review of important vocabulary terms and characteristics of circles that we can later on apply to solving our sample problems.

Are you ready to get started?

 

Figure 01: What are the key properties of circles?

 

What is the Area of a Circle?

In order to understand what the area of a circle actually represents, it’s important that you are familiar with some key properties of circles:

  • Every circle has a center point.

  • The circumference of a circle (also known as the edge or perimeter of the circle) is the outer boundary of the circle. All points on the edge of a circle are equidistant from the center point.

  • The diameter of a circle is a line that passes through the center point and touches two points on the edge of the circle. Any diameter will cut the circle in half.

  • The radius of a circle is a line segment that starts at the center and ends on the edge of the circle. Any radius will always be equal to half the length of the circle’s diameter.

All of these key properties are highlighted for Circle P in Figure 01 above.

The area of a circle is the amount of two-dimensional space that is contained inside of the circle. Area is measured in square units.

What does this mean in real-world terms? If you wanted to figure out the size of the region contained within a particular circle, you would simply have to calculate its area. For example, if you wanted to tile the bottom of a circular swimming pool, knowing the size of the pool’s area would help you to determine how many square units of tile you would need to cover the entire region.

This real-world example is illustrated in Figure 02 below.

 

Figure 02: Real-world application of finding the area of a circle. (Image: Mashup Math MJ)

 

While many math problems involving circles with simply include two-dimensional diagrams of circles, the idea is still the same. The key takeaway here is that the area of a circle is the size of the region encompassed within the circle. The larger the circle, the larger its area will be. And, since circles are two-dimensional figures (i.e. they are flat), the area will be measured in square units.

Figure 03 illustrates this concept and what it means when we say that the area of a circle will always be expressed in square units.

Notice that many of the squares are “cut off” and incomplete, so we should expect to have answers that include decimals rather than whole numbers (we will see this more when we get to the practice problems later on).

For now, just make sure that you are comfortable with the key properties of circles (especially the difference between the radius of a circle and the diameter of a circle) and what the area of a circle represents before moving onto the next section where we will learn the area of a circle formula.

 

Figure 03: What is the area of a circle?

 

Area of a Circle Formula: A=πr²

You can find the area of a circle by using the area of a circle formula: A=πr² (where r is the length of the radius of the circle).

Whenever you are calculating a circle’s area using the area of a circle formula, you can use pi (π) or an approximation of pi (π≈3.14).

Before moving forward, make sure that you remember the area of a circle formula and that the r in the formula represents the length of the radius of whatever circle whose area you are trying to find. The diagram in Figure 04 below can serve as a reference for remembering the area of a circle formula. You may also wish to write the formula down before moving onto the practice problems later on in this guide.

 

Figure 04: The area of a circle formula.

 

For example, if we wanted to find the area of a circle with a radius of 3 cm, we could use the area of a circle formula to determine the approximate amount of square inches that make up the region inside of the circle.

In this case,

  • A=πr²

  • r=3

So, to find the area of this circle, we would simply have to substitute r with 3 and then evaluate to find the area as follows:

  • A=πr²

  • A=π(3)²

  • A=π9 or A=9π

  • A=28.27433388…

  • A=28.3 cm²

As shown above, we know that area equals π times 3². Since we know that 3²=9, we can now say that area equals π times 9 (or 9π). Finally, we can put 9π into a calculator to get 28.27433388….

In this guide, we will find area to the nearest tenth of a square unit, so we can approximate our final answer as A=28.3 cm².

These steps are illustrated in Figure 05 below.

 

Figure 05: How to use the area of a circle formula to find the area of a circle.

 

Next, let’s move onto working on a few more examples of how to find the area of a circle with a given radius or a given diameter. Then, we will take a look at how to find the area of a half circle.


How to Find Area of a Circle with a Radius

In the example shown in Figure 05 above, we found the area of a circle with a radius of 3cm using the area of a circle formula A=πr².

Whenever you have to find the area of a circle where the radius is given, you can simply input the given value of r (the radius of the circle) into the area of a circle formula and evaluate to find the value of the area.

And we will go ahead and do exactly that in the example below.

Example #1: Find the Area of a Circle with Radius 7 mm

For this example, we have to find the area of a circle with a radius of 7 mm.

We can easily find the area of using the area of a circle formula, A=πr², by substituting r with 7 as follows:

  • A=πr²

  • A=π(7)²

  • A=π49 or A=49π

  • A=153.93804…

  • A=153.9 mm²

That’s all that there is to it! Whenever you know the length of the radius of a circle, r, you can input that value into the area of a circle formula to figure out its area.

The steps for solving this first example are illustrated in Figure 06 below. Make sure that you are comfortable with finding the area of a circle with a given radius using the area of a circle formula before moving onto the next section where we will learn how to find area of a circle with diameter.

 

Figure 06: How to Find the Area of a Circle Step-by-Step

 

How to Find Area of a Circle with Diameter

While students often have a relatively easy time with finding the area of a circle when they know the value of its radius (as shown in the previous section), they can sometimes get confused when they have to find the area of a circle with diameter given instead of radius.

However, you can still use the area of a circle formula, A=πr², to find the area of a circle with a diameter, but with one small extra step.

Remember that the radius of a circle is a line segment that starts at the center and ends on the edge of the circle and that any radius will always be equal to half the length of the circle’s diameter.

In other words, the diameter of a circle is equal to twice the length of the circle’s radius. Or, vice versa, the radius of a circle is equal to half the length of a circle’s diameter.

So, if you know the diameter of a circle, then you can simply divide it by two to find its radius.

For example, consider a circle with a diameter of 10 feet. This means that the circle has a radius of 5 feet. And, if you know the radius of a circle, then you can use the area of a circle formula to calculate its area.

Let’s go ahead and learn how to find the area of a circle with diameter with this in mind by working through the examples below.

 

Figure 07: The radius of any circle is equal to one half the length of its diameter.

 

Example #2: Find the Area of a Circle with Diameter 24 yd

For this example, we have to find the area of a circle with a diameter of 24 yards.

Just like the previous examples, we can use the area of a circle formula, A=πr², to find the area of the circle in question.

However, we can’t use the formula just yet because we only know the length of the diameter and not the radius. So, in order to find the value of the radius, r, we have to first have to divide the diameter by two to find the value of r as follows:

  • r=d/2

  • r=24/2

  • r=12

So, if the circle has a diameter of 24 yards, then it also has a radius of 12 yards (because 24÷2=12).

Now that we know the value of r (r=12 in this case), we can substitute it into the area of a circle formula and solve as follows:

  • A=πr²

  • A=π(12)²

  • A=π144 or A=144π

  • A=452.3893421…

  • A=452.4 yd²

And that is how to find area of a circle with diameter!

Just remember that you can always determine the length of a circle’s radius by dividing its diameter by two. And once you know a circle’s radius, you can use the formula A=πr² to find its area.

 

Figure 08: How to find the area of a circle with diameter by using the area of a circle formula.

 

Next, let’s look at another example of how to find area of a circle with diameter in a real world context.

Example #3: Find the Area of a Circle with Diameter 57 m.

For this next example, we want to find the area of a circular zen garden with a diameter of 57 meters.

Again, we can solve this problem by using the area of a circle formula A=πr².

Since we know that the circle has a diameter of d=57, we can use that given information to find the length of the radius as follows:

Figure 09: Find the area of a circular zen garden with 57 m diameter.

  • r=d/2

  • r=57/2

  • r=28.5

Remember that we can always find the length of a circle’s radius simply by dividing its diameter by two. In this case, 57/2=28.5 so we can say that r=28.5 (note that it’s totally fine to have a radius that is a decimal value).

So, if the circular zen garden has a diameter of 57 m, then it also has a radius of 28.5 m (because 57÷2=28.5).

Now that we know that r=28.5, we can substitute it into the area of a circle formula and solve as follows:

  • A=πr²

  • A=π(28.5)²

  • A=π812.25 or A=812.25π

  • A=2,551.758633…

  • A=2,551.8 m²

We can now conclude that the circular zen garden has an approximate area of 2,551.8 square meters and we have solved the problem!

 

Figure 10: Real world example of how to find the area of a circle with diameter given.

 

Now, let’s move onto the final section where we will learn how to find the area of a half circle.


How to Find Area of a Half Circle

Now that you know how to find the area of a circle with a given radius or diameter using the area of a formula circle, you are ready to learn how to find the area of a half circle (also known as a semicircle).

What is a half circle? Simply put, a half circle is a regular circle that has been cut in half along the diameter.

And, notably, the area of a half circle is exactly one half the area of a regular circle with the same diameter.

The differences between a circle and a half circle are illustrated in Figure 11 below.

 

Figure 11: What is the difference between the area of a circle and the area of a half circle?

 

With this difference in mind, we will have to use a modified version of the area of a circle formula to find the area of a half circle.

  • Area of a Circle Formula: A=πr²

  • Area of a Half Circle Formula: A=(πr²)/2

Notice that the area of a half circle formula is the same as the area of a circle formula except that the result is being dividing by 2. This should make sense because the area of a half circle is half the size of the area of a regular circle with the same diameter.

 

Figure 12: The area of a half circle formula is A=(πr²)/2

 

Now that you know the area of a half circle formula, let’s learn how to find the area of a half circle step-by-step.

Example #4: Find the Area of a Half Circle with Diameter 22 cm

For this example, we have to find the area of a half circle with a diameter of 22 cm.

To solve this problem, we can use the area of a half circle formula: A=(πr²)/2

Let’s start by figuring out the value of r (the radius of the half circle), which can determine by cutting the diameter in half as follows:

  • r=d/2

  • r=22/2

  • r=11

Now that we know the value of r (r=11 in this example), we can substitute it into the area of a half circle formula and solve as follows:

  • A=(πr²)/2

  • A=(π(11)²)/2

  • A=(121π)/2

  • A=190.0663555…

  • A=190.1 cm²

And that is how to find area of a half circle using the area of a half circle formula.

All of the steps for solving this example are illustrated in Figure 13 below.

 

Figure 13: How to Find Area of a Half Circle

 

Example #5: Find the Area of a Half Circle

For this final example, let’s learn how to find the area of a circle (namely a half circle) in a real-world scenario.

In this case, we have a circular shaped table with a surface that is half-covered with tile and half-covered with hardwood and we want to find the area of the portion that is covered in hardwood.

If we know that the table has a diameter of 13 feet, we can find the length of the radius use the area of a half circle formula to solve the problem as follows:

  • r=d/2

  • r=13/2

  • r=6.5

Now that we know that r=6.5, we can substitute it into the area of a half circle formula and solve as follows:

  • A=(πr²)/2

  • A=(π(6.5)²)/2

  • A=(42.25π)/2

  • A=66.36614481…

  • A=66.4 ft²

We can now conclude that the hardwood half of the table has an area of 66.4 square feet and we are finished!

 

Figure 14: How to find the area of a half circle with a diameter of 13 feet.

 

How to Find the Area of a Circle: Conclusion

Learning how to find the area of a circle or a half circle is an important and useful math skill.

You can find the area of a circle by using the formula A=πr² where r is the length of the circle’s radius.

If you only know the length of a circle’s diameter, you can simply divide it by two (since d=r/2) to find the value of its radius and then use the formula to find the circle’s area.

You can find the area of a half circle by using the formula A=(πr²)/2 where r is the length of the circle’s radius. This formula is derived from the regular area of a circle formula since the area of a half circle is equal to half the area of a regular circle with the same diameter.

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