Volume of a Orthohedron - Examples, Exercises and Solutions

The formula to calculate the volume of a cuboid is:

height*length*width

We replace the data in the formula:  

3*5*7

7*5 = 35

35*3 = 105

Volume formula for a rectangular prism:

Volume = length X width X height

Therefore, first we will place the data we are given into the formula:

45 = 2.5*4*X

We divide both sides of the equation by 2.5:

18=4*X

And now we divide both sides of the equation by 4:

4.5 = X

The formula for the volume of a box is height*length*width

In the specific question, we are given the volume and the height,

and we are looking for the area of the base,

As you will remember, the area is length * width

If we replace all the data in the formula, we see that:

4 * the area of the base = 80

Therefore, if we divide by 4 we see that

Area of the base = 20

To calculate the volume of a cuboid, as we mentioned, we need the length, width, and height.

It is important to note that in the exercise we are given the height and the base area of the cuboid.

The base area is actually the area multiplied by the length. That is, it is the data that contains the two pieces of information we are missing.

Therefore, we can calculate the area by height * base area

15*3 = 45

This is the solution!

We know the area of DBHF and also the length of HF

We will substitute into the formula in order to find BF, let's call the side BF as X:

$4\times x=20$

We'll divide both sides by 4:

$x=5$

Therefore, BF equals 5

Now we can calculate the volume of the box:

$8\times4\times5=32\times5=160$

Based on the given data, we can argue that:

$BD=HF=2$

We know the area of ABCD and also the length of DB

We'll substitute in the formula to find CD, let's call the side CD as X:

$2\times x=12$

We'll divide both sides by 2:

$x=6$

Therefore, CD equals 6

Now we can calculate the volume of the box:

$6\times2\times3=12\times3=36$

Since we are given the area of rectangle CAEG and length AE, we can find GE:

Let's denote GE as X and substitute the data in the rectangle area formula:

$3\times x=15$

Let's divide both sides by 3:

$x=5$

Therefore GE equals 5

Since we are given the area of rectangle ABFE and length AE, we can find EF:

Let's denote EF as Y and substitute the data in the rectangle area formula:

$3\times y=24$

Let's divide both sides by 3:

$y=8$

Therefore EF equals 8

Now we can calculate the volume of the box:

$3\times5\times8=15\times8=120$

We use a formula to calculate the volume: height times width times length.

We rewrite the exercise using the existing data:

$21\times(14+30x)\times15=$

We use the distributive property to simplify the parentheses.

We multiply 21 by each of the terms in parentheses:

$(21\times14+21\times30x)\times15=$

We solve the multiplication exercise in parentheses:

$(294+630x)\times15=$

We use the distributive property again.

We multiply 15 by each of the terms in parentheses:

$294\times15+630x\times15=$

We solve each of the exercises in parentheses to find the volume:

$4,410+9,450x$