Question: An official playing field (including end zones) for the Indoor Football League has a length 38 yd…

An official playing field (including end zones) for the Indoor Football League has a length 38 yd longer than its width. The perimeter of the rectangular field is 160 yd. Find the length and width of the field.

The length of the rectangle is \(\boxed{\quad}\) and the width of the rectangle is \(\boxed{\quad}\).

Solution

The problem states that an official playing field for the Indoor Football League has a length 38 yards longer than its width, with a perimeter of 160 yards. We are to find the length and width of the rectangle. Let the width of the rectangle be \( x \). Then the length of the rectangle is \( x + 38 \). The formula for the perimeter of a rectangle is: \[ P = 2 \times (\text{length} + \text{width}) \] Substitute the given values into the formula: \[ 160 = 2 \times (x + x + 38) \] Simplify the equation: \[ 160 = 2 \times (2x + 38) \] \[ 160 = 4x + 76 \] Subtract 76 from both sides: \[ 160 - 76 = 4x \] \[ 84 = 4x \] Divide both sides by 4: \[ x = 21 \] The width of the rectangle is 21 yards. The length is: \[ x + 38 = 21 + 38 = 59 \] The length of the rectangle is 59 yards.