# [SOLVED] Working out the volume of a cuboid

• Oct 13th 2009, 09:43 AM
dave1022
[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 $x$cm. The distance from one corner of the cuboid to the furthest corner is $y$cm. What is the total surface area of the cube in terms of $x$ and $y$.

This is what I've got so far:

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

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

I've tried to re-arrange those equations to get the surface area in terms of $x$ and $y$, 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?
• Oct 13th 2009, 10:52 AM
aman_cc
Quote:

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 $x$cm. The distance from one corner of the cuboid to the furthest corner is $y$cm. What is the total surface area of the cube in terms of $x$ and $y$.

This is what I've got so far:

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

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

I've tried to re-arrange those equations to get the surface area in terms of $x$ and $y$, 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
$(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ca)$
• Oct 13th 2009, 11:44 AM
dave1022
Thanks for the hint!
Now I've got:

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

Therefore:

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

Therefore:

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

Thanks for the hint :D