Questions: The width of a rectangular slab of concrete is 2 m less than the length. The area is 35 m^2. Part: 0 / 3 Part 1 of 3 (a) What are the dimensions of the rectangle? The length of the slab is m.

Solution Steps

To find the dimensions of the rectangular slab, we need to set up an equation using the given information. Let the length of the rectangle be \( l \) meters. Then, the width will be \( l - 2 \) meters. The area of the rectangle is given by the product of its length and width, which is 35 square meters. We can set up the equation \( l \times (l - 2) = 35 \) and solve for \( l \).

1. Let the length of the rectangle be \( l \).
2. The width of the rectangle is \( l - 2 \).
3. Set up the equation for the area: \( l \times (l - 2) = 35 \).
4. Solve the quadratic equation to find the value of \( l \).

Let the length of the rectangular slab be \( l \) meters. The width is then given by \( w = l - 2 \) meters.

The area \( A \) of the rectangle is given by the formula: \[ A = l \times w \] Substituting the expression for width, we have: \[ A = l \times (l - 2) = 35 \]

Expanding the equation: \[ l^2 - 2l - 35 = 0 \] Using the quadratic formula, we find the solutions for \( l \): \[ l = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-35)}}{2 \cdot 1} \] This simplifies to: \[ l = \frac{2 \pm \sqrt{4 + 140}}{2} = \frac{2 \pm \sqrt{144}}{2} = \frac{2 \pm 12}{2} \] Thus, the solutions are: \[ l = \frac{14}{2} = 7 \quad \text{and} \quad l = \frac{-10}{2} = -5 \] Since length cannot be negative, we take \( l = 7 \).

Now, substituting \( l \) back to find the width: \[ w = l - 2 = 7 - 2 = 5 \]

Final Answer