Question

the area of a rectangle is 70 m², and the length of the rectangle is 11 m less than three times the width. find the dimensions of the rectangle. length : \boxed{} m width : \boxed{} m

Explanation:

Step1: Define variables

Let the width of the rectangle be \( w \) meters. Then the length \( l \) is \( 3w - 11 \) meters (since length is 11 m less than three times the width).

Step2: Use area formula

The area of a rectangle is given by \( A = l \times w \). We know the area \( A = 70 \, \text{m}^2 \), so substitute \( l = 3w - 11 \) into the area formula:

$$ (3w - 11) \times w = 70 $$

Expand the left side:

$$ 3w^2 - 11w = 70 $$

Rearrange into standard quadratic form:

$$ 3w^2 - 11w - 70 = 0 $$

Step3: Solve the quadratic equation

We can solve \( 3w^2 - 11w - 70 = 0 \) using the quadratic formula \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = -11 \), and \( c = -70 \).

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

$$ D = (-11)^2 - 4 \times 3 \times (-70) = 121 + 840 = 961 $$

Then, find the square root of the discriminant: \( \sqrt{961} = 31 \).

Now, substitute into the quadratic formula:

$$ w = \frac{-(-11) \pm 31}{2 \times 3} = \frac{11 \pm 31}{6} $$

We have two solutions:

1. \( w = \frac{11 + 31}{6} = \frac{42}{6} = 7 \)
2. \( w = \frac{11 - 31}{6} = \frac{-20}{6} = -\frac{10}{3} \) (discard since width can't be negative)

Step4: Find the length

Now that we have \( w = 7 \) meters, substitute into the length formula \( l = 3w - 11 \):

$$ l = 3 \times 7 - 11 = 21 - 11 = 10 $$

Answer:

Length: \( 10 \) m, Width: \( 7 \) m