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
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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**: \( 10.24 \, \text{m} \)
