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