Question

the length of a rectangle is 1 m more than twice the width, and the area of the rectangle is $45\\ m^2$. 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 \( 2w + 1 \) meters (since length is 1 m more than twice the width).

Step2: Use area formula

The area of a rectangle is \( A = l \times w \). We know \( A = 45 \, \text{m}^2 \), so substitute \( l = 2w + 1 \) into the area formula:

$$ (2w + 1)w = 45 $$

Expand the left side:

$$ 2w^2 + w = 45 $$

Rearrange into standard quadratic form:

$$ 2w^2 + w - 45 = 0 $$

Step3: Solve the quadratic equation

We can solve \( 2w^2 + w - 45 = 0 \) using the quadratic formula \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = 1 \), and \( c = -45 \).

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

$$ D = 1^2 - 4(2)(-45) = 1 + 360 = 361 $$

Then, find the square root of the discriminant: \( \sqrt{361} = 19 \)

Now, substitute into the quadratic formula:

$$ w = \frac{-1 \pm 19}{2(2)} $$

We have two solutions:

1. \( w = \frac{-1 + 19}{4} = \frac{18}{4} = 4.5 \)
2. \( w = \frac{-1 - 19}{4} = \frac{-20}{4} = -5 \)

Since width cannot be negative, we discard \( w = -5 \). So the width \( w = 4.5 \) meters.

Step4: Find the length

Substitute \( w = 4.5 \) into the length formula \( l = 2w + 1 \):

$$ l = 2(4.5) + 1 = 9 + 1 = 10 $$

Answer:

Length: \( 10 \) m
Width: \( 4.5 \) m