# Volume of Simple 3D Shapes  Rating

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The authors Eugene Lee

## Basics on the topicVolume of Simple 3D Shapes

After this lesson, you will be able to solve the volume of simple three-dimensional shapes.

The lesson begins by teaching you three examples of simple three-dimensional shapes. It leads you to learn how to find the volume of these shapes by calculating the area of their base and multiplying the result by their height. It concludes with showing the different formulas for calculating the volume of simple three-dimensional shapes.

Learn about the volume of simple three-dimensional shapes by helping the Mad Hatter, a world-renowned milliner, create the perfect headpiece for the queen!

This video introduces new concepts, notation, and vocabulary such as volume (the amount of space that a three-dimensional object occupies); rectangular box (a three-dimensional shape with a rectangular base); triangular prism (a three-dimensional shape with a triangular base); cylinder (a three-dimensional shape with a circular base); area (the amount of space that a two-dimensional object occupies); height (distance from the top to the bottom of an object); and cubic units (the unit of measure used for measuring the volume of an object).

Before watching this video, you should already be familiar with rectangles, triangles, circles, dimensions of two-dimensional shapes like length and width, radius of a circle, units of measurement, and Product Law of Exponents.

After watching this video, you will be prepared to learn to solve more real-world and mathematical problems involving volumes of three-dimensional objects.

Common Core Standard(s) in focus: 7.G.B.4 and 7.G.B.6 A video intended for math students in the 7th grade Recommended for students who are 12-13 years old

## Volume of Simple 3D Shapes exercise

Would you like to apply the knowledge you’ve learned? You can review and practice it with the tasks for the video Volume of Simple 3D Shapes.
• ### Decide which statements about volume are true.

Hints

Here are the formulas for the base area and volume of a rectangular prism.

If you divide a rectangle with a diagonal, you get two congruent triangles. Therefore, each triangle has half the area of the rectangle.

Solution

Calculating the volume of a prism always follows the same general idea:

• First, determine the area of the base: the units will be $in^2$ or $ft^2$ or ...
• Then, multiply this area by the height of the prism, because the base is repeated that many times. The resulting units will be $in^3$ or $ft^3$ or ...
$V_{prism}=Area_{base}\cdot height_{prism}$

There are several area formulas that are used quite often and should be memorized:

• circle: $A=\pi\cdot r^2$ where $r$ is the radius.
• triangle: $A=\frac12 base_{\Delta}\cdot height_{\Delta}$.
• rectangle: $A=lw$ where $l$ is the length and $w$ is the width.
• ### Determine the volume of a cylinder.

Hints

The concept for the calculation of the volume of a prism is always the same:

• Determine the area of the base.
• Multiply this area by the height of the prism.

Because $pi$ is greater than $1$, the result using an approximation of $\pi$ must be greater than the factor by which you multiply $\pi$.

Solution

We use the volume formula

$V=Area_{\circ}\cdot height_{box}$

• Putting the radius $r=21~in$ in the area formula we get $Area_{\circ}=\pi(25~in)^2=625\pi~in^2$.
• Now we multiply this result by $height_{box}=51~in$. This results in $V=31,875\pi~in^3$.
• Putting $\pi\approx 3.14$ in this result gives us $V\approx 31,875(3.14)~in^3=100,087.5~in^3$.
• ### Identify formulas for the 3-D objects and their cross sections.

Hints

Look at the labeling of the 3-D shapes.

You have to assign an area as well as a volume formula to each shape.

The volume formula must match the area formula.

Solution

Here you can see all the formulas for area and volume of rectangular prisms, triangular prisms, and cylinders.

Rectangular prism

• The base area is given by $area_{\square}=l\cdot w$.
• We multiply this by the remaining length $h$ to get the volume $V=l\cdot w\cdot h$.
Triangular prism

• The base area is given by $area_{\Delta}=base_{\Delta}\cdot height_{\Delta}$.
• We multiply this by the height of the box $h$ to get the volume $V=base_{\Delta}\cdot height_{\Delta}\cdot h$.
Cylinder

• The base area is given by $area_{\circ}=\pi\cdot r^2$.
• We multiply this by the height of the box $h$ to get $V=\pi\cdot r^2\cdot h$.
• ### Solve these real world problems.

Hints

For the volume of a rectangular prism just multiply the side lengths

$V=l\cdot w\cdot h$.

Pay attention when rounding numbers:

• $3.1415$ rounded to two decimal places results in $3.14$.
• $2.71828$ rounded to two decimal places results in $2.72$.
When rounding to two decimal places, if the thousandths place is greater than or equal you have to round up. Otherwise you have to round down.

Use the circumference formula for a circle:

$C=2\cdot \pi\cdot r$.

Substitute the value for the circumference and solve for the radius.

The volume of a cylinder is given by

$V=\pi\cdot r^2\cdot h$

where $r$ is the radius and $h$ the height of the cylinder.

Solution

Pool 1 For this pool we just have to multiply the side lengths:

$V=(48~in)^2\cdot (12~in)=27,648~in^3$

Pool 2 Since this pool is a cylindrical prism, we have to use the volume formula for cylinders

$V=\pi\cdot r^2\cdot h$

where $r$ is the radius and $h$ the height of the cylinder. Plugging in the known values we get

$V=\pi\cdot (36~in)^2\cdot (8~in)= 10,368 \pi~in^3\approx10,368 \cdot 3.14~in^3= 32,555.52 ~in^3$

Thus we can conclude that the square prism holds less water than the cylinder.



A standard $8.5$ by $11$ inch sheet of paper

Cylinder A is formed by bringing the two longer sides around together.

So we can determine the radius with $8.5=2\cdot \pi\cdot r$. Dividing by $2\cdot pi\approx 2\cdot 3.14=6.28$ we get $r\approx 1.35~in$.

Now we can put this in the volume formula for cylinders we used above:

$V=\pi\cdot (1.35~in)^2\cdot (11~in)= 20.0475 \pi~in^3\approx20.0475 \cdot 3.14~in^3\approx 62.95 ~in^3$

Cylinder B is formed by bringing the two shorter sides together.

Here we determine the radius by using $11=2\cdot \pi\cdot r$. Dividing by $2\cdot pi\approx 2\cdot 3.14=6.28$ we get $r\approx 1.75~in$.

Now we can put this in the volume formula for cylinders from above:

$V=\pi\cdot (1.75~in)^2\cdot (8.5~in)= 26.03125 \pi~in^3\approx 26.03125 \cdot 3.14~in^3\approx 81.74 ~in^3$

• ### Identify the shapes.

Hints

Each prism has a base and a top face which are congruent.

A pyramid has a base and a peak.

Each corner of the base, for example a square or a rectangle, is connected with this peak.

Here you see a sphere. This is not a prism.

Solution

What are the common features of prisms?

• The bottom and top faces are congruent.
• They consist of the bottom and top faces, plus several rectangles or a curved surface
Six faces which are rectangles could indicate a rectangular prism or as a special case a cube:

A cube has only squares as faces. All of the six faces are congruent.

A prism with a circle as the bottom and top face is called a cylinder.

Last, we have to investigate a prism with triangles as the bottom and top faces: This is a triangular prism.

• ### Find mistakes in the volume calculations.

Hints

Don't forget about units! Each time you multiply an area ($in^2$) by a length ($in$), you also have to multiply the units.

The formula for the area of a circle is $A=\pi\cdot r^2$.

There are seven mistakes in total.

Solution

Mr. Joke's cylinder:

The radius is $r=1.4~in$ and the height $h=2.5~in$.

So we get for the volume

$V=\pi\cdot (1.4~in)^2\cdot (2.5~in)=4.9\pi~in^3\approx 15.4~in^3$

(in the problem above, the radius and height were switched!)

Don't forget about the units: it's $in^3$.

$~$

Paul's Lunchbox:

• First we multiply all side lengths to get the volume
$\quad~~~V=(6~in)\cdot (4~in)\cdot (2~in)=48~in^3$.

• Because we know the volume of one slice of bread, $12~in^3$, we just have to divide $48~in^3$ by $12~in^3$ to get the number of slices of bread: It's $4$, not $6$!
$~$

Carol's Carrot Cake:

• The formula for a triangle's area is given by $A=\frac12\cdot base_{\Delta}\cdot height_{\Delta}=\frac12(3~in)\cdot(3~in)=4.5~in^2$
It's not multiplied by $\frac{1}{3}$, and the prism height of $6~in$ is not used. Instead, we use the triangle height.

• This area must be multiplied by the height $h=6~in$ to get $V=(4.5~in^2)\cdot (6~in)=27~in^3$
Watch the units!