If the perimeters of two similar triangles are in the ratio of 4 : 7, and the sum of the areas is 195 cm², then what is $\frac{1}{3}$ of the difference between the areas (in cm²) of the two triangles?

The correct answer is Option 2: 33.

We are given that two triangles are similar and the ratio of their perimeters is 4 : 7.

Let the perimeters of the two similar triangles be $P_1$ and $P_2$, and their respective areas be $A_1$ and $A_2$.

By the property of similar triangles, the ratio of their areas is equal to the square of the ratio of their perimeters (or corresponding sides):

$\frac{A_1}{A_2}={\left(\frac{P_1}{P_2}\right)}^2$

Given that $\frac{P_1}{P_2}=\frac{4}{7}$, we substitute this into the formula:

$\frac{A_1}{A_2}={\left(\frac{4}{7}\right)}^2=\frac{16}{49}$

Thus, we can represent the areas as:

$A_1=16x$
$A_2=49x$

where $x$ is a common constant ratio.

We are given that the sum of the areas of the two triangles is 195 cm²:

$A_1+A_2=195$

Substitute the expressions in terms of $x$:

$16x+49x=195$

$65x=195$

$x=\frac{195}{65}=3$

Now, let's find the areas of the two triangles:

$A_1=16\times 3=48{\text{cm}}^2$
$A_2=49\times 3=147{\text{cm}}^2$

Next, we find the difference between the areas of the two triangles:

$\text{Difference}=A_2-A_1=147-48=99{\text{cm}}^2$

Finally, we calculate $\frac{1}{3}$ of the difference between the areas:

$\frac{1}{3}\times 99=33{\text{cm}}^2$

Therefore, $\frac{1}{3}$ of the difference between the areas of the two triangles is 33 cm².