# Thread: finding volume with algebra

1. ## finding volume with algebra

PUZZLING PRISM:
The areas of three faces of a rectangular prism are 63 cm2, 56 cm2 and 72 cm2. What is the volume of the prism?

What I know:

Volume= LxWxH
Area = LxW

2. Originally Posted by puppy_wish

PUZZLING PRISM:
The areas of three faces of a rectangular prism are 63 cm2, 56 cm2 and 72 cm2. What is the volume of the prism?

What I know:

Volume= LxWxH
Area = LxW

You know that,
$xy=63$
$xz=56$
$yz=72$
Multiply them together,
$xyxzyz=63\cdot 56\cdot 72=254016$
$x^2y^2z^2=(xyz)^2=V^2=254016$
Thus,
$V=\sqrt{254016}=504$

3. Hello, puppy_wish!

This can be solved with "normal" algebra
. . but there is a cute solution . . .

The areas of three faces of a rectangular prism are 63 cm², 56 cm² and 72 cm².
What is the volume of the prism?

Let the length, width, height be: $L,\,W,\,H.$

We are told that: . $\begin{array}{ccc} LW \,=\,63 \\ WH\,=\,56 \\ LW\,=\,72\end{array}$

Multiply the three equations: . $(LW)(WH)(LH) \:=\:(63)(56)(72)$

and we have: . $L^2W^2H^2 \:=\:254,016\quad\Rightarrow\quad(LWH)^2\:=\:254,0 16$

Therefore: . $LWH \:=\:\sqrt{254,016} \:=\:504\text{ cm}^2$

I like this solution . . . I hope you do.
We answered the questi0on without finding $L,\,W$ and $H.$

4. Originally Posted by Soroban
Hello, puppy_wish!

This can be solved with "normal" algebra
. . but there is a cute solution . . .

Let the length, width, height be: $L,\,W,\,H.$

We are told that: . $\begin{array}{ccc} LW \,=\,63 \\ WH\,=\,56 \\ LW\,=\,72\end{array}$

Multiply the three equations: . $(LW)(WH)(LH) \:=\63)(56)(72)" alt="(LW)(WH)(LH) \:=\63)(56)(72)" />

and we have: . $L^2W^2H^2 \:=\:254,016\quad\Rightarrow\quad(LWH)^2\:=\:254,0 16$

Therefore: . $LWH \:=\:\sqrt{254,016} \:=\:504\text{ cm}^2$

I like this solution . . . I hope you do.
We answered the questi0on without finding $L,\,W$ and $H.$

wow thank you both of you. I really appreciate it