Math Snap

STEP 1

What is this asking?
We need to find the volume of four different 3D shapes: a trapezoidal prism, a pentagonal pyramid, a cylinder, and another pentagonal pyramid!
Watch out!
Don't mix up the formulas for different shapes!
Also, be careful with the units – we've got meters, miles, feet, and kilometers!

STEP 2

1. Trapezoidal Prism Volume
2. Pentagonal Pyramid Volume
3. Cylinder Volume
4. Pentagonal Pyramid Volume

STEP 3

Alright, for the trapezoidal prism, the formula for the volume is $V = \frac{1}{2} (b_1 + b_2) \cdot h_1 \cdot h_2$, where $b_1$ and $b_2$ are the lengths of the parallel bases of the trapezoid, $h_1$ is the height of the trapezoid, and $h_2$ is the height or depth of the prism.
It's like finding the area of the trapezoid and then multiplying by how deep the prism goes!

STEP 4

Let's plug in the values!
We have $b_1 = 4$ m, $b_2 = 8$ m, $h_1 = 3$ m, and $h_2 = 4.3$ m.
So, $V = \frac{1}{2} (4 + 8) \cdot 3 \cdot 4.3$.

STEP 5

First, add those bases: $4 + 8 = 12$.
Now we have $V = \frac{1}{2} \cdot 12 \cdot 3 \cdot 4.3$.

STEP 6

Multiply $1/2$ by $12$ to get $6$.
Now, $V = 6 \cdot 3 \cdot 4.3$.

STEP 7

Multiply $6$ and $3$ to get $18$.
Finally, $V = 18 \cdot 4.3 = 77.4 \text{ m}^3$.
So, the volume of the trapezoidal prism is $\mathbf{77.4} \text{ }\mathbf{m^3}$.

STEP 8

The volume of a pyramid is $V = \frac{1}{3} \cdot B \cdot h$, where $B$ is the area of the base and $h$ is the height of the pyramid.
Here, the base is a pentagon, but they've already given us the area!

STEP 9

We're given that the area of the pentagonal base is $5 \cdot 5 = 25$ mi$^2$ and the height is $h = 3$ mi.
Plugging these values into the formula, we get $V = \frac{1}{3} \cdot 25 \cdot 3$.

STEP 10

Multiplying $\frac{1}{3}$ by $3$ gives us $1$, so $V = 25 \cdot 1 = 25$ mi$^3$.
The volume of this pentagonal pyramid is $\mathbf{25} \text{ }\mathbf{mi^3}$.

STEP 11

The volume of a cylinder is $V = \pi r^2 h$, where $r$ is the radius and $h$ is the height.
We're given the diameter, which is $22$ ft, so the radius is half of that, $r = 11$ ft.
The height is given as $h = 10$ ft.

STEP 12

Let's plug in the values: $V = \pi \cdot 11^2 \cdot 10$.

STEP 13

Squaring the radius gives $11^2 = 121$, so $V = \pi \cdot 121 \cdot 10$.

STEP 14

Multiplying $121$ by $10$ gives $1210$, so $V = \pi \cdot 1210$.
Using $\pi \approx 3.14$, we get $V \approx 3.14 \cdot 1210 = 3799.4$ ft$^3$.
The volume of the cylinder is approximately $\mathbf{3799.4} \text{ }\mathbf{ft^3}$.

STEP 15

Just like the previous pentagonal pyramid, the volume is $V = \frac{1}{3}Bh$.
We're given that the area of the base is $9 \cdot 10 = 90$ km$^2$ and the height is $6$ km.

STEP 16

Plugging in the values, we get $V = \frac{1}{3} \cdot 90 \cdot 6$.

STEP 17

Multiplying $\frac{1}{3}$ by $90$ gives $30$, so $V = 30 \cdot 6$.

STEP 18

Finally, $V = 180$ km$^3$.
The volume of this pentagonal pyramid is $\mathbf{180} \text{ }\mathbf{km^3}$.

7) $77.4 \text{ m}^3$
8) $25 \text{ mi}^3$
9) $3799.4 \text{ ft}^3$
10) $180 \text{ km}^3$