1. **State the problem:** We need to find the perimeter of a triangle with side lengths $\frac{3}{4}$ m, $\frac{3}{5}$ m, and $\frac{3}{4}$ m. 2. **Formula for perimeter:** The perimeter $P$ of a triangle is the sum of the lengths of all its sides: $$P = a + b + c$$ where $a$, $b$, and $c$ are the side lengths. 3. **Substitute the given side lengths:** $$P = \frac{3}{4} + \frac{3}{5} + \frac{3}{4}$$ 4. **Add the fractions:** First, add the two $\frac{3}{4}$ terms: $$\frac{3}{4} + \frac{3}{4} = \frac{6}{4} = \frac{3}{2}$$ Now add $\frac{3}{2}$ and $\frac{3}{5}$: Find a common denominator, which is 10: $$\frac{3}{2} = \frac{15}{10}, \quad \frac{3}{5} = \frac{6}{10}$$ So, $$P = \frac{15}{10} + \frac{6}{10} = \frac{21}{10}$$ 5. **Convert to mixed number:** $$\frac{21}{10} = 2 \frac{1}{10}$$ **Final answer:** The perimeter of the triangle is $\boxed{\frac{21}{10}}$ meters or $\boxed{2 \frac{1}{10}}$ meters.