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# 10.7: Heights of Cylinders Given Surface Area

Difficulty Level: At Grade Created by: CK-12
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Practice Heights of Cylinders Given Surface Area

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What if you know the surface area of a cylinder, but don't know the height? Look at this dilemma.

Candice is going to use modge podge to decorate the outside of a cylindrical canister. She wants to make it into a decorative pencil holder as a gift for her Mom. Modge podge is a glue - like substance that helps to adhere tissue paper or other decorative pieces of paper to an object.

Candice's canister has a radius of 3 inches and has a surface area of 207.3 in2\begin{align*}207.3 \ in^2\end{align*}.

What is the height of the canister?

Use this Concept to learn about how to figure out the height of a cylinder given the radius and surface area.

### Guidance

Sometimes we may be given the surface area of a cylinder and need to find its height. We simply fill the information we know into the surface area formula. This time, instead of solving for SA\begin{align*}SA\end{align*}, we solve for h\begin{align*}h\end{align*}, the height of the cylinder.

Let’s look at how we can apply this information.

The surface area of a cylinder with a radius of 3 inches is 78π\begin{align*}78 \pi\end{align*} square inches. What is the height of the cylinder?

We know that the radius of the cylinder is 3 inches. We also know that the surface area of the cylinder is 78π\begin{align*}78 \pi\end{align*} square inches. Sometimes a problem will include pi in the amount because it is more precise. We’ll see that this also makes our calculations easier. All we have to do is put 78π\begin{align*}78 \pi\end{align*} into the formula in place of SA\begin{align*}SA\end{align*}. Then we can use the formula to solve for the height, h\begin{align*}h\end{align*}.

SA78π78π78π78π18π60π60π÷6π10 in.=2πr2+2πrh=2π(32)+2π(3)h=2π(9)+6πh=18π+6πh=6πhSubtract 18π from both sides.=6πh=hDivide both sides by 6π.=h\begin{align*}SA & = 2 \pi r^2 + 2 \pi rh\\ 78 \pi & = 2 \pi (3^2) + 2 \pi (3) h\\ 78 \pi & = 2 \pi (9) + 6 \pi h\\ 78 \pi & = 18 \pi + 6 \pi h\\ 78 \pi - 18 \pi & = 6 \pi h \qquad \qquad \qquad \text{Subtract} \ 18 \pi \ \text{from both sides.}\\ 60 \pi & = 6 \pi h\\ 60 \pi \div 6 \pi & = h \qquad \qquad \quad \qquad \text{Divide both sides by} \ 6 \pi.\\ 10 \ in. & = h \end{align*}

We used the formula to find that a cylinder with a radius of 3 inches and a surface area of 78π\begin{align*}78 \pi\end{align*} has a height of 10 inches.

We can check our work by putting the height into the formula and solving for surface area.

SASASASA=2π(32)+2π(3)(10)=2π(9)+2π(30)=18π+60π=78π\begin{align*}SA & = 2 \pi (3^2) + 2 \pi (3) (10)\\ SA & = 2 \pi (9) + 2 \pi (30)\\ SA & = 18 \pi + 60 \pi\\ SA & = 78 \pi\end{align*}

This is the surface area given in the problem, so our answer is correct. Let’s look at another example.

Now try a few of these on your own.

Find the height in each problem.

#### Example A

SA=48π, r=3 in\begin{align*}SA = 48 \pi, \ r = 3 \ in\end{align*}

Solution:2 in\begin{align*}2 \ in\end{align*}

#### Example B

SA=60π, r=2 in\begin{align*}SA = 60 \pi, \ r = 2 \ in\end{align*}

Solution:5 in\begin{align*}5 \ in\end{align*}

#### Example C

SA=80π, r=4 m\begin{align*}SA = 80 \pi, \ r = 4 \ m\end{align*}

Solution: 2 m\begin{align*}2 \ m\end{align*}

Here is the original problem once again.

Candice is going to use modge podge to decorate the outside of a cylindrical canister. She wants to make it into a decorative pencil holder as a gift for her Mom. Modge podge is a glue - like substance that helps to adhere tissue paper or other decorative pieces of paper to an object.

Candice's canister has a radius of 3 inches and has a surface area of 207.24 in2\begin{align*}207.24 \ in^2\end{align*}.

What is the height of the canister?

To figure this out, let's use the formula.

SA66π66π66π66π18π66π66π÷6π8 in.=2πr2+2πrh=2π(32)+2π(3)h=2π(9)+6πh=18π+6πh=6πhSubtract 18π from both sides.=6πh=hDivide both sides by 6π.=h\begin{align*}SA & = 2 \pi r^2 + 2 \pi rh\\ 66 \pi & = 2 \pi (3^2) + 2 \pi (3) h\\ 66 \pi & = 2 \pi (9) + 6 \pi h\\ 66 \pi & = 18 \pi + 6 \pi h\\ 66 \pi - 18 \pi & = 6 \pi h \qquad \qquad \qquad \text{Subtract} \ 18 \pi \ \text{from both sides.}\\ 66 \pi & = 6 \pi h\\ 66 \pi \div 6 \pi & = h \qquad \qquad \quad \qquad \text{Divide both sides by} \ 6 \pi.\\ 8 \ in. & = h \end{align*}

The height of the cylinder is 8 in\begin{align*}8 \ in\end{align*}.

### Vocabulary

Here are the vocabulary words in this Concept.

Cylinder
a three-dimensional figure with two circular bases.
Surface Area
the measurement of the outside of a three-dimensional figure.
Net
a two-dimensional representation of a three-dimensional figure.

### Guided Practice

Here is one for you to try on your own.

Mrs. Javitz bought a container of oatmeal in the shape of the cylinder. If the container has a radius of 5 cm and a surface area of 170π\begin{align*}170 \pi\end{align*}, what is its height?

Once again, we have been given the surface area as a function of pi. That’s no problem we simply put the whole number, 170π\begin{align*}170 \pi\end{align*}, in for SA\begin{align*}SA\end{align*} in the formula. We also know that the container has a radius of 5 centimeters, so we can use the formula to solve for h\begin{align*}h\end{align*}, the height of the container.

SA170π170π170π170π50π120π120π÷10π12 cm=2πr2+2πrh=2π(52)+2π(5)h=2π(25)+10πh=50π+10πh=10πhSubtract 50π from both sides.=10πh=h  Divide both sides by 10π.=h\begin{align*}SA & = 2 \pi r^2 + 2 \pi rh\\ 170 \pi & = 2 \pi (5^2) + 2 \pi (5) h\\ 170 \pi & = 2 \pi (25) + 10 \pi h\\ 170 \pi & = 50 \pi + 10 \pi h\\ 170 \pi - 50 \pi & = 10 \pi h \quad \qquad \qquad \text{Subtract} \ 50 \pi \ \text{from both sides.}\\ 120 \pi & = 10 \pi h\\ 120 \pi \div 10 \pi & = h \qquad \qquad \qquad \ \ \text{Divide both sides by} \ 10 \pi.\\ 12 \ cm & = h\end{align*}

The oatmeal container must have a height of 12 centimeters.

### Video Review

Here is a video for review.

### Practice

Directions: Find the height of each cylinder given the surface area and one other dimension.

1. r=5 in\begin{align*}r = 5 \ in\end{align*}

SA=376.8 in2\begin{align*}SA = 376.8 \ in^2\end{align*}

2. r=6 in\begin{align*}r = 6 \ in\end{align*}

SA=527.52 in2\begin{align*}SA = 527.52 \ in^2\end{align*}

3. r=5.5 in\begin{align*}r = 5.5 \ in\end{align*}

SA=336.8255 in2\begin{align*}SA = 336.8255 \ in^2\end{align*}

4. r=4 cm\begin{align*}r = 4 \ cm\end{align*}

SA=251.2 cm2\begin{align*}SA = 251.2 \ cm^2\end{align*}

5. r=5 m\begin{align*}r = 5 \ m\end{align*}

SA=439.6 m2\begin{align*}SA = 439.6 \ m^2\end{align*}

6. r=2 m\begin{align*}r = 2 \ m\end{align*}

SA=87.92 cm2\begin{align*}SA = 87.92 \ cm^2\end{align*}

7. r=2 cm\begin{align*}r = 2 \ cm\end{align*}

SA=75.35 cm2\begin{align*}SA = 75.35 \ cm^2\end{align*}

8. d=4 m\begin{align*}d = 4 \ m\end{align*}

SA=125.6 m2\begin{align*}SA = 125.6 \ m^2\end{align*}

9. d=8 cm\begin{align*}d = 8 \ cm\end{align*}

SA=351.68 cm2\begin{align*}SA = 351.68 \ cm^2\end{align*}

10. d=6 cm\begin{align*}d = 6 \ cm\end{align*}

SA=282.6 cm2\begin{align*}SA = 282.6 \ cm^2\end{align*}

11. r=3 cm\begin{align*}r = 3 \ cm\end{align*}

\begin{align*}SA = 226.08 \ cm^2\end{align*}

12. \begin{align*}r = 14 \ cm\end{align*}

\begin{align*}SA = 2637.6 \ cm^2\end{align*}

13. \begin{align*}r = 18 \ cm\end{align*}

\begin{align*}SA = 4295.53 \ cm^2\end{align*}

14. \begin{align*}r = 12 \ cm\end{align*}

\begin{align*}SA = 1959.36 \ cm^2\end{align*}

15. \begin{align*}r = 13 \ cm\end{align*}

\begin{align*}SA = 3347.24 \ cm^2\end{align*}

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