Volume of a Orthohedron: Using variables

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:

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

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:

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

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:

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

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:

$4,410+9,450x$