Question

the area of a rectangle is 65 m², and the length of the rectangle is 3 m less than twice the width. find the dimensions of the rectangle. length : m width : m

Explanation:

Step1: Define Variables

Let the width of the rectangle be \( w \) meters. Then the length \( l \) is \( 2w - 3 \) meters (since length is 3 m less than twice the width).

Step2: Use Area Formula

The area of a rectangle is \( A = l \times w \). We know \( A = 65 \, \text{m}^2 \), so substitute \( l = 2w - 3 \) into the formula:

$$ (2w - 3)w = 65 $$

Expand the left side:

$$ 2w^2 - 3w = 65 $$

Rearrange into standard quadratic form:

$$ 2w^2 - 3w - 65 = 0 $$

Step3: Solve Quadratic Equation

We can solve \( 2w^2 - 3w - 65 = 0 \) using the quadratic formula \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = -3 \), and \( c = -65 \).

First, calculate the discriminant \( D = b^2 - 4ac \):

$$ D = (-3)^2 - 4(2)(-65) = 9 + 520 = 529 $$

Then, find \( w \):

$$ w = \frac{-(-3) \pm \sqrt{529}}{2(2)} = \frac{3 \pm 23}{4} $$

We have two solutions:

$$ w_1 = \frac{3 + 23}{4} = \frac{26}{4} = 6.5 \quad \text{(discard negative solution as width can't be negative)} $$

$$ w_2 = \frac{3 - 23}{4} = \frac{-20}{4} = -5 \quad \text{(invalid, so we take } w = 6.5 \text{)} $$

Step4: Find Length

Now that \( w = 6.5 \) m, find length \( l = 2w - 3 \):

$$ l = 2(6.5) - 3 = 13 - 3 = 10 \, \text{m} $$

Answer:

Length: \( 10 \) m, Width: \( 6.5 \) m