EXERCISE 11 (a) The perimeter of a rectangle is 44 m and the area is \( 117 \mathrm{~m}^{2} \). Determine the dimensions of the rectangle.

To find the dimensions of the rectangle, we can use the formulas for perimeter and area. The perimeter \(P\) of a rectangle is given by \(P = 2(l + w)\) and the area \(A\) is given by \(A = l \times w\), where \(l\) is the length and \(w\) is the width. Given \(P = 44\) m, we have: \[ l + w = 22 \] (1) And since \(A = 117 \, \text{m}^2\): \[ l \cdot w = 117 \] (2) From equation (1), we can express \(w\) in terms of \(l\): \[ w = 22 - l \] Now substitute \(w\) into equation (2): \[ l(22 - l) = 117 \] Expanding gives: \[ 22l - l^2 = 117 \] Rearranging into standard quadratic form: \[ l^2 - 22l + 117 = 0 \] Now, we can use the quadratic formula \(l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=1\), \(b=-22\), and \(c=117\): \[ l = \frac{22 \pm \sqrt{(-22)^2 - 4 \cdot 1 \cdot 117}}{2 \cdot 1} \] \[ l = \frac{22 \pm \sqrt{484 - 468}}{2} \] \[ l = \frac{22 \pm \sqrt{16}}{2} \] \[ l = \frac{22 \pm 4}{2} \] Calculating both possibilities: 1. \(l = \frac{26}{2} = 13\) 2. \(l = \frac{18}{2} = 9\) Thus, the dimensions \(l\) and \(w\) can be \(13\) m and \(9\) m, respectively. Therefore, the dimensions of the rectangle are \(13 \, \text{m}\) by \(9 \, \text{m}\).