Volume of a Orthohedron: Using variables

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:

$4,410+9,450x$