Urgent! Help! Similar Solids!

• Jan 7th 2007, 02:59 PM
sarahgasoline
Urgent! Help! Similar Solids!
Two solids are similar, with volumes of 54cm^3 and 16cm^3. If the larger solid has a surface area of 27cm^2, what is the surface area of the smaller solid?

So its a proportion...

16cm^3 X
_______ = ______
54cm^3 27cm^2

But how do i solve it? Cross multiply? do i have to turn them into cubed instead of squared?
• Jan 7th 2007, 03:53 PM
Quick
Quote:

Originally Posted by sarahgasoline
Two solids are similar, with volumes of 54cm^3 and 16cm^3. If the larger solid has a surface area of 27cm^2, what is the surface area of the smaller solid?

So its a proportion...

16cm^3 X
_______ = ______
54cm^3 27cm^2

But how do i solve it? Cross multiply? do i have to turn them into cubed instead of squared?

Let's see what happens if we have a 2x2x2 cube (volume=8, s.a.=24) and a 4x4x4 cube (volume=64, s.a.=92)

does $\frac{8}{64}=\frac{24}{96}$?

However, does: $\frac{8}{24}=\frac{64}{96}$

in fact, I think you need to know what kind of object it is...

but you should wait for someone else's opinion.
• Jan 8th 2007, 12:48 AM
topsquark
Quote:

Originally Posted by sarahgasoline
Two solids are similar, with volumes of 54cm^3 and 16cm^3. If the larger solid has a surface area of 27cm^2, what is the surface area of the smaller solid?

So its a proportion...

16cm^3 X
_______ = ______
54cm^3 27cm^2

But how do i solve it? Cross multiply? do i have to turn them into cubed instead of squared?

Say there is a size factor of x between the lengths of the respective sides $L_1$ and $L_2$ of the solid. That is to say: $L_2 = x L_1$. (This ratio obviously will hold for any chosen linear measurement between the two solids. This is the definition of similar solids.) Then the volumes will be in the cube of that ratio:
$V_2 = x^3V_1$

In this case:
$54 \, cm^3 = x^3(16 \, cm^3)$

$x^3 = \frac{54 \, cm^3}{16 \, cm^3} = \frac{27}{8}$

$x = \frac{3}{2}$

Now, surface areas will be in the square of the ratio:
$SA_2 = x^2SA_1$

So
$27 \, cm^2 = \left ( \frac{3}{2} \right ) ^2SA_1$

$27 \, cm^2 = \frac{9}{4} \cdot SA_1$

$SA_1 = \frac{4}{9} ( 27 \, cm^2 )$

$SA_1 = 12 \, cm^2$

-Dan
• Jan 8th 2007, 01:56 AM
ticbol
Quote:

Originally Posted by sarahgasoline
Two solids are similar, with volumes of 54cm^3 and 16cm^3. If the larger solid has a surface area of 27cm^2, what is the surface area of the smaller solid?

So its a proportion...

16cm^3 X
_______ = ______
54cm^3 27cm^2

But how do i solve it? Cross multiply? do i have to turn them into cubed instead of squared?

Here is one way.

You cannot equate ratios that are not in the same power. Example, you cannot equate a ratio of cubes to a ratio of squares. They have to be in the same power. A ratio of cubes to another ratio of cubes, ....

(16 cu.cm)/(54 cu.cm) is a ratio of cubes.
(x sq.cm)/(27 sq.cm) is a ratio of squares.
The two cannot be equal.

So, either (a) reduce the (16/54) into a ratio of squares, or (b) raise the (x/27) into a ratio of cubes.

a) All in ratios of cubes:
(16/54) = (x/27)^(3/2)
In the righthand side, we took first the square roots of x and 27, then we cubed those roots. Now both ratios are in cubes.
To continue,
Clear the radical, raise both sides to their (2/3) powers,
(16/54)^(2/3) = (x/27)
x = 27*[(16/54)^(2/3)]