Let's denote the length of the rectangle as \(\( L \)\) and the width as \(\( W \)\). We are given the perimeter (\(\( P \)\)) and the area (\(\( A \)\)) of the rectangle:
\(\[ P = 8 \text{ m} \]\)
\(\[ A = 2 \text{ m}^2 \]\)
The perimeter of a rectangle is calculated by the formula \(\( P = 2L + 2W \)\). Since we are given that the perimeter is 8 m, we can write:
\(\[ 2L + 2W = 8 \]\)
Dividing both sides by 2, simplifies the equation:
\(\[ L + W = 4 \]\) (Equation 1)
The area of a rectangle is calculated by the formula \(\( A = L \times W \)\). Given that the area is 2 m², we have:
\(\[ L \times W = 2 \]\) (Equation 2)
To find the length and the width, we can use the system of equations formed by Equation 1 and Equation 2. From Equation 1, we can express \(\( W \)\) in terms of \(\( L \)\):
\(\[ W = 4 - L \]\)
Substitute \(\( W \)\) into Equation 2 to find \(\( L \)\):
\(\[ L \times (4 - L) = 2 \]\)
\(\[ 4L - L^2 = 2 \]\)
Rearranging the equation, we bring it to a standard quadratic form:
\(\[ L^2 - 4L + 2 = 0 \]\)
To solve for \(\( L \)\), we will apply the quadratic formula which is:
\(\[ L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]\)
For our quadratic equation, \(\( a = 1 \)\), \(\( b = -4 \)\), and \(\( c = 2 \)\).
First, let's calculate the discriminant (\(\( \Delta \)\)):
\(\[ \Delta = b^2 - 4ac \]\)
\(\[ \Delta = (-4)^2 - 4 \cdot 1 \cdot 2 \]\)
\(\[ \Delta = 16 - 8 \]\)
\(\[ \Delta = 8 \]\)
Now, let's use the quadratic formula:
\(\[ L = \frac{-(-4) \pm \sqrt{8}}{2 \cdot 1} \]\)
\(\[ L = \frac{4 \pm \sqrt{8}}{2} \]\)
\(\[ L = \frac{4 \pm 2\sqrt{2}}{2} \]\)
\(\[ L = 2 \pm \sqrt{2} \]\)
We get two solutions for \(\( L \)\):
\(\[ L_1 = 2 + \sqrt{2} \]\)
\(\[ L_2 = 2 - \sqrt{2} \]\)
Since the length of a rectangle must be positive, we discard \(\( L_2 \)\) as it will give a negative width when we substitute it back into Equation 1. Therefore, we use the positive solution:
\(\[ L_1 = 2 + \sqrt{2} \approx 2 + 1.4 \approx 3.4 \text{ m} \]\)
Now let's calculate the width using \(\( L_1 \)\):
\(\[ W = 4 - L \]\)
\(\[ W = 4 - (2 + \sqrt{2}) \]\)
\(\[ W = 4 - 2 - \sqrt{2} \]\)
\(\[ W = 2 - \sqrt{2} \approx 2 - 1.4 \approx 0.6 \text{ m} \]\)
So, to the nearest tenth of a metre, the length of the rectangle is approximately 3.4 m, and the width is approximately 0.6 m.
