Question

if the area of a base of a rectangular solid is tripled, what is the percent increase in the volume?

Answer 1

here is a trick to use. since it does not matter what the base is make it a number.
say you have a cube with edge length 2 one and so the area of the base is 2^2=4 and the volume is 2^3=8.
then you triple the area of the base to 12. now the side of edge of the cube has length
\[\sqrt{12}=2\sqrt{3}\] and the volume is
\[\sqrt{12}^3=(2\sqrt{3})^3=8\sqrt{3}^3=72\sqrt{3}\]
you want the percent increase so divide
\[\frac{72\sqrt{3}}{8}=9\sqrt{3}\]

Answer 2

why you want to write this number as a percent is beyond me, but if you triple the are of the base the volume increases by a factor of
\[9\sqrt{3}\] or
\[9^{\frac{3}{2}}\]

Answer 3

but the question is asking for the percent increase. it's either 200, 300, 600 or 900.

Answer 4

900

Answer 5

the answer key says 200 for some reason.

Answer 6

then i guess i am wrong...

Answer 7

no in fact i am right. take a cube with side edge 1, so base has side 1 and volume is 1. triple the area of the base. it is 3, the edge is
\[\sqrt{3}\] and the volume is
\[\sqrt{3}^3=9\sqrt{3}\] that it the increase, from 1 to
\[9\sqrt{3}\]

Answer 8

The question is asking about a rectangle, not a cube. So can anyone kindly explain the correct answer?