The area of a rectangle is \( 52 \mathrm{~m}^{2} \), and the length of the rectangle is 5 m less than thice the width. Find the dimensions of the rectangle

To solve for the dimensions of the rectangle, we start with the formulas for area and dimensions. Let the width be \( w \) meters. Then the length \( l \) can be expressed as: \[ l = 3w - 5 \] Given that the area \( A \) is: \[ A = l \times w = 52 \, \text{m}^2 \] Substituting the expression for length: \[ (3w - 5)w = 52 \] This simplifies to: \[ 3w^2 - 5w - 52 = 0 \] Now we can use the quadratic formula, \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = -5 \), and \( c = -52 \): \[ w = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 3 \cdot (-52)}}{2 \cdot 3} \] \[ w = \frac{5 \pm \sqrt{25 + 624}}{6} \] \[ w = \frac{5 \pm \sqrt{649}}{6} \] Calculating \( \sqrt{649} \), we get approximately \( 25.5 \): \[ w = \frac{5 \pm 25.5}{6} \] This gives two possible values for \( w \): 1. \( w = \frac{30.5}{6} \approx 5.08 \, \text{m} \) 2. \( w = \frac{-20.5}{6} \) (which is not feasible since width cannot be negative) Now, substituting \( w \) back to find \( l \): \[ l = 3(5.08) - 5 \approx 15.24 - 5 = 10.24 \, \text{m} \] Thus, the dimensions of the rectangle are approximately: **Width**: \( 5.08 \, \text{m} \) **Length**: \( 10.24 \, \text{m} \)