Question 235253
<font face="Garamond" size="+2">


Let *[tex \Large s] represent the measure of one edge of a cube.  The volume is then *[tex \Large s^3] and the surface area is *[tex \Large 6s^2].  If you triple the measure of the edges, then each edge would measure *[tex \Large 3s], so the volume would be *[tex \Large (3s)^3\ =\ 27s^3] which is 27 times larger than the original cube.  The surface area of the larger cube would be *[tex \Large 6(3s)^2 \ =\ 6\,\cdot\,9s^2] which is 9 times larger than the original cube.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>