1.3 Pquations \[ \quad \begin{array}{l}\text { Question 11, 1.3.25 } \\ \text { Part } 1 \text { of } 2 \\ \text { A rectangular athletic field is twice as long as it is wide. If the perimeter of the athletic field is } 216 \text { yards, what are } \\ \text { The width is } \square \text { yards. }\end{array} \] To.5 of 15

To find the width of the rectangular athletic field, we can set up some equations based on the information given. Let's denote the width as \( w \) yards. Since the length is twice the width, we express the length as \( 2w \). The formula for the perimeter \( P \) of a rectangle is given by \( P = 2(\text{length} + \text{width}) \). In this case, we know the perimeter is 216 yards, so we can set up the equation: \[ 216 = 2(2w + w) \] \[ 216 = 2(3w) \] \[ 216 = 6w \] Dividing both sides by 6 gives us \( w = 36 \) yards. Now we can check our work. If the width is 36 yards, then the length is \( 2 \times 36 = 72 \) yards. The perimeter would be \( 2(72 + 36) = 216 \) yards, confirming our solution.