Length : \( \square \mathrm{m} \) The area of a rectangle is \( 70 \mathrm{~m}^{2} \), and the length of the rectangle is 11 m less than three times th dimensions of the rectangle. Width : \( \square \mathrm{m} \)

Let the width be W (in meters) and the length be L (in meters). According to the problem, the rectangle’s area is 70 m² and its length is 11 m less than three times its width. This gives us the following equations:   1) L × W = 70   2) L = 3W − 11 Substitute equation (2) into equation (1):   (3W − 11) × W = 70   3W² − 11W − 70 = 0 Now, we solve the quadratic equation 3W² − 11W − 70 = 0. The quadratic formula is:   W = [−b ± √(b² − 4ac)] / (2a) Here, a = 3, b = −11, and c = −70. First, calculate the discriminant:   Δ = (−11)² − 4(3)(−70)   Δ = 121 + 840   Δ = 961 Since √961 = 31, we have:   W = [11 ± 31] / (2 × 3) This gives two potential solutions:   W = (11 + 31) / 6 = 42 / 6 = 7 m   W = (11 − 31) / 6 = (−20) / 6 = −10/3 m Because the width must be positive, we take W = 7 m. Now substitute back to find L:   L = 3(7) − 11 = 21 − 11 = 10 m So the dimensions of the rectangle are:   Length: 10 m   Width: 7 m