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### Course: High school geometry>Unit 9

Lesson 3: Volume and surface area

# Volume of composite figures

It's time to mix and match!  Now that we can find volume in several basic shapes, let's combine them to see how we model more interesting and useful real figures.

## Composite volume

It's time to mix and match! Now that we can find volume in several basic shapes, let's combine them to see how we model more interesting and useful real figures.

## Subtracting to find volume

A cylinder has been drilled out of this block to allow a rod to pass through. All measurements are in millimeters. Let's find the remaining volume of the block.

### Subdividing the figure

The figures is composed of a rectangular prism with a cylinder removed from it.

### Finding the separate volumes

What is the volume of the rectangular prism?
${\text{mm}}^{3}$

What is the volume of the cylinder?
${\text{mm}}^{3}$

### Subtracting to find the total volume

Let's subtract to find the volume.
What is the remaining volume of the block?

Let's consider the volume of a tent with the following dimensions. All measurements are in meters. The base of the tent is a rectangle.

### Subdividing the figure

It looks a little like a triangular prism, but it sticks out too far at the bottom. We can split this figure into the triangular prism and two halves of a pyramid.

### Calculating separate volumes

Let's start with the volume of the triangular prism. All prisms have a volume of $\left(\text{base area}\right)\left(\text{height}\right)$.
What is the area of the triangular base?
${\text{m}}^{2}$

What is the volume of the triangular prism?
${\text{m}}^{3}$

Now we can find the volume of the pyramid. Let’s combine the two half pyramids and find the volume of that whole pyramid.
All pyramids have a volume of $\frac{1}{3}\left(\text{base area}\right)\left(\text{height}\right)$.
What is the area of the rectangular base of the whole pyramid?
${\text{m}}^{2}$

What is the volume of the whole pyramid?
${\text{m}}^{3}$

### Adding to find the total volume

Finally, we can add the volumes of the triangular prism and the pyramid to find the total volume of the tent.
What is the volume of the tent?
${\text{m}}^{3}$

## Challenge problem

A sharpened pencil is shaped like a right circular cone attached to a cylinder, each with the same radius. The pencil, including the point, is $190\phantom{\rule{0.167em}{0ex}}\text{mm}$ long, and the unsharpened base has an area of of $40\phantom{\rule{0.167em}{0ex}}{\text{mm}}^{2}$. The volume of the pencil is $7,040\phantom{\rule{0.167em}{0ex}}{\text{mm}}^{3}$.
We're going to solve for the length $p$ of the sharpened part of the pencil.

### Expressing the volumes in terms of the unknown

Write an expression for the volume of the cone in terms of $p$.

Write an expression for the height of the cylinder, in terms of $p$.

Write an expression for the volume of the cylinder, in terms of $p$.

Write an expression for the entire volume of the pencil, in terms of $p$.

### Solving for the unknown

Now we can set your expression for the volume of the pencil equal to the numeric volume, $7,040\phantom{\rule{0.167em}{0ex}}{\text{mm}}^{3}$ and solve for $p$.
What is the length $p$ of the sharpened portion of the pencil?
$\text{mm}$

## Want to join the conversation?

• i am so lost and confused
• bro me to
• this fr makes no since
• it's hard in some parts
• A company makes spherical tanks that store chemicals. Their standard model has a diameter of 666 meters, but a customer needs a larger tank whose capacity is 2.52.52, point, 5 times the volume of the standard model.
What is the approximate diameter of this larger tank?
• So you have a sphere with a diameter of 6, and you want to find the diameter of a sphere that has a volume 2.5 times as greater. The least mentally taxing way to do this is the calculate the volume of the first sphere, then multiply by 2.5, and then apply the formula in reverse to find the radius and then diameter of the second sphere.
d = 2r, so the radius of the first sphere is 3m
V = 4/3 * pi * r^3
V = 4/3 * pi * (3 * 3 * 3)
V = 36pi
Now to get the volume of the tank that is 2.5 times as large, multiply by 2.5 and get 90pi m^3
V = 4/3 * pi * r^3
r = cbrt( V / (4/3 * pi))
r = 4.07
Multiply by 2 to get a new diameter of 8.14 meters.

If you realize that if you change the radius by a certain factor, the Volume changes by the factor cubed. So, since you're changing the volume by 2.5, you basically change the radius by cbrt(2.5).
3 * cbrt(2.5) will give you the new radius then.
• I like how some people on here are just copy-pasting math assignment/test questions.
• I did the problems but I can't seem to understand why and how I got some of the answers.
I do better with the videos. :|
It's frustrating. >:(
• Why can't I say that the volume of the cylinder is the volume of the entire pencil minus the volume of the cone? volume of cylinder + volume of cone = volume of pencil
--> volume of cylinder = volume of pencil - volume of cone
--> volume of cyliner = 7040 - 40p / 3
• You can. The way you posed is just a re-formatted version of the solution they offered. (i.e they offered [V(cyl)+V(cone)=V(pen)], and you just subtracted V(cone) from both sides. Either way, you'd get the same answer.