# Thread: [SOLVED] Working out the volume of a cuboid

1. ## [SOLVED] Working out the volume of a cuboid

Here is the information I've been given:

The sum of the lengths of the 12 edges of a cuboid is $\displaystyle x$cm. The distance from one corner of the cuboid to the furthest corner is $\displaystyle y$cm. What is the total surface area of the cube in terms of $\displaystyle x$ and $\displaystyle y$.

This is what I've got so far:

Call the three lengths on the cuboid $\displaystyle a, b, c$.

The total surface area = $\displaystyle 2ab+2ac+2bc$
As $\displaystyle y$ is the longest diagonal, $\displaystyle y^2=a^2+b^2+c^2$
And finally, as $\displaystyle x$ is the total lengths, $\displaystyle x=4a+4b+4c$

I've tried to re-arrange those equations to get the surface area in terms of $\displaystyle x$ and $\displaystyle y$, but to no avail.

All help appreciated,
David

EDIT: I've just realised that I've put this in the wrong sub-forum. Please could one of the moderators move this to the appropriate sub-forum?

2. Originally Posted by dave1022
Here is the information I've been given:

The sum of the lengths of the 12 edges of a cuboid is $\displaystyle x$cm. The distance from one corner of the cuboid to the furthest corner is $\displaystyle y$cm. What is the total surface area of the cube in terms of $\displaystyle x$ and $\displaystyle y$.

This is what I've got so far:

Call the three lengths on the cuboid $\displaystyle a, b, c$.

The total surface area = $\displaystyle 2ab+2ac+2bc$
As $\displaystyle y$ is the longest diagonal, $\displaystyle y^2=a^2+b^2+c^2$
And finally, as $\displaystyle x$ is the total lengths, $\displaystyle x=4a+4b+4c$

I've tried to re-arrange those equations to get the surface area in terms of $\displaystyle x$ and $\displaystyle y$, but to no avail.

All help appreciated,
David

EDIT: I've just realised that I've put this in the wrong sub-forum. Please could one of the moderators move this to the appropriate sub-forum?
Hint
$\displaystyle (a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ca)$

3. ## Thanks for the hint!

Now I've got:

$\displaystyle (x^2)/16=a^2+b^2+c^2+2(ab+bc+ca)$

Therefore:

$\displaystyle (x^2)/16=y^2+2(ab+bc+ca)$

Therefore:

Surface area$\displaystyle =((x^2)/16)-y^2$

Thanks for the hint