1. **Stating the problem:** We have an equilateral triangle with side length $4a + 2$ millimeters and a square with side length $2a + 7$ millimeters. We want to compare their perimeters. 2. **Formulas for perimeters:** - Perimeter of an equilateral triangle: $$P_{triangle} = 3 \times \text{side length}$$ - Perimeter of a square: $$P_{square} = 4 \times \text{side length}$$ 3. **Calculate the perimeter of the triangle:** $$P_{triangle} = 3(4a + 2) = 3 \times 4a + 3 \times 2 = 12a + 6$$ 4. **Calculate the perimeter of the square:** $$P_{square} = 4(2a + 7) = 4 \times 2a + 4 \times 7 = 8a + 28$$ 5. **Compare the perimeters:** We want to find when the perimeters are equal or which is larger. Set them equal: $$12a + 6 = 8a + 28$$ 6. **Solve for $a$:** Subtract $8a$ from both sides: $$12a + 6 - \cancel{8a} = \cancel{8a} + 28 - 8a \Rightarrow 4a + 6 = 28$$ Subtract 6 from both sides: $$4a + 6 - 6 = 28 - 6 \Rightarrow 4a = 22$$ Divide both sides by 4: $$\frac{\cancel{4}a}{\cancel{4}} = \frac{22}{4} \Rightarrow a = \frac{22}{4} = 5.5$$ 7. **Interpretation:** - If $a = 5.5$, the perimeters are equal. - For $a < 5.5$, the square's perimeter is larger. - For $a > 5.5$, the triangle's perimeter is larger. **Final answer:** $$a = 5.5$$ is the value where the perimeters are equal.