To solve this problem, let's carefully break down the information given and proceed step-by-step:
1. We know that the area of triangle [tex]\( S \)[/tex] is defined as 1.
2. Triangle [tex]\( R \)[/tex] is 3 times the area of triangle [tex]\( S \)[/tex]. So, the area of triangle [tex]\( R \)[/tex] is:
[tex]\[
\text{Area of } R = 3 \times \text{Area of } S = 3 \times 1 = 3
\][/tex]
3. Triangle [tex]\( S \)[/tex] is 3 times the area of triangle [tex]\( T \)[/tex]. So, the area of triangle [tex]\( T \)[/tex] is:
[tex]\[
\text{Area of } T = \frac{\text{Area of } S}{3} = \frac{1}{3}
\][/tex]
Now, we need to find the sum of the areas of triangle [tex]\( R \)[/tex], triangle [tex]\( S \)[/tex], and triangle [tex]\( T \)[/tex]:
[tex]\[
\text{Sum of areas} = \text{Area of } R + \text{Area of } S + \text{Area of } T
\][/tex]
Substituting the values we have:
[tex]\[
\text{Sum of areas} = 3 + 1 + \frac{1}{3}
\][/tex]
To add these values, it is helpful to convert all terms to have a common denominator:
[tex]\[
3 = 3 \quad (or \, 3 = \frac{9}{3})
\][/tex]
[tex]\[
1 = 1 \quad (or \, 1 = \frac{3}{3})
\][/tex]
[tex]\[
\frac{1}{3} = \frac{1}{3}
\][/tex]
Now, adding these fractions together:
[tex]\[
\frac{9}{3} + \frac{3}{3} + \frac{1}{3} = \frac{9 + 3 + 1}{3} = \frac{13}{3} = 4 \frac{1}{3}
\][/tex]
So, the sum of the areas of the three triangles is:
[tex]\[
4 \frac{1}{3}
\][/tex]
Therefore, the correct answer is:
[tex]\[
\boxed{4 \frac{1}{3}}
\][/tex]
This corresponds to option [tex]\( c \)[/tex].
