Answer :
To find the dimensions of the rectangle, we start by defining the variables involved and using the given information.
1. Let's denote the width of the rectangle as [tex]\( w \)[/tex] meters.
2. The length of the rectangle is given as 3 meters more than twice the width. Therefore, the length can be represented as [tex]\( 2w + 3 \)[/tex] meters.
Next, we know that the area of the rectangle is given to be 54 square meters. The area of a rectangle is calculated by multiplying its length by its width. Therefore, we set up the equation:
[tex]\[ \text{Area} = \text{Length} \times \text{Width} \][/tex]
Substituting the given expressions for the area and length:
[tex]\[ 54 = (2w + 3) \times w \][/tex]
We now solve this equation step-by-step to find [tex]\( w \)[/tex]:
1. Distribute [tex]\( w \)[/tex] to both terms in the parenthesis on the right-hand side of the equation:
[tex]\[ 54 = 2w^2 + 3w \][/tex]
2. Re-arrange the equation to set it to 0:
[tex]\[ 2w^2 + 3w - 54 = 0 \][/tex]
3. This is a quadratic equation in the form [tex]\( ax^2 + bx + c = 0 \)[/tex]. We solve for [tex]\( w \)[/tex] using the quadratic formula [tex]\( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 2 \)[/tex], [tex]\( b = 3 \)[/tex], and [tex]\( c = -54 \)[/tex].
[tex]\[ w = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-54)}}{2 \cdot 2} \][/tex]
[tex]\[ w = \frac{-3 \pm \sqrt{9 + 432}}{4} \][/tex]
[tex]\[ w = \frac{-3 \pm \sqrt{441}}{4} \][/tex]
[tex]\[ w = \frac{-3 \pm 21}{4} \][/tex]
4. This gives us two potential solutions:
[tex]\[ w = \frac{-3 + 21}{4} = \frac{18}{4} = 4.5 \][/tex]
And
[tex]\[ w = \frac{-3 - 21}{4} = \frac{-24}{4} = -6 \][/tex]
Since a width cannot be negative, we discard [tex]\( w = -6 \)[/tex] and accept [tex]\( w = 4.5 \)[/tex] meters.
Now, we substitute [tex]\( w \)[/tex] back into the expression for the length:
[tex]\[ \text{Length} = 2w + 3 \][/tex]
[tex]\[ \text{Length} = 2(4.5) + 3 = 9 + 3 = 12 \, \text{meters} \][/tex]
Thus, the dimensions of the rectangle are:
- Width: [tex]\( 4.5 \)[/tex] meters
- Length: [tex]\( 12 \)[/tex] meters
So, the results are:
Length: [tex]\( 12 \)[/tex] meters
Width: [tex]\( 4.5 \)[/tex] meters
1. Let's denote the width of the rectangle as [tex]\( w \)[/tex] meters.
2. The length of the rectangle is given as 3 meters more than twice the width. Therefore, the length can be represented as [tex]\( 2w + 3 \)[/tex] meters.
Next, we know that the area of the rectangle is given to be 54 square meters. The area of a rectangle is calculated by multiplying its length by its width. Therefore, we set up the equation:
[tex]\[ \text{Area} = \text{Length} \times \text{Width} \][/tex]
Substituting the given expressions for the area and length:
[tex]\[ 54 = (2w + 3) \times w \][/tex]
We now solve this equation step-by-step to find [tex]\( w \)[/tex]:
1. Distribute [tex]\( w \)[/tex] to both terms in the parenthesis on the right-hand side of the equation:
[tex]\[ 54 = 2w^2 + 3w \][/tex]
2. Re-arrange the equation to set it to 0:
[tex]\[ 2w^2 + 3w - 54 = 0 \][/tex]
3. This is a quadratic equation in the form [tex]\( ax^2 + bx + c = 0 \)[/tex]. We solve for [tex]\( w \)[/tex] using the quadratic formula [tex]\( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 2 \)[/tex], [tex]\( b = 3 \)[/tex], and [tex]\( c = -54 \)[/tex].
[tex]\[ w = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-54)}}{2 \cdot 2} \][/tex]
[tex]\[ w = \frac{-3 \pm \sqrt{9 + 432}}{4} \][/tex]
[tex]\[ w = \frac{-3 \pm \sqrt{441}}{4} \][/tex]
[tex]\[ w = \frac{-3 \pm 21}{4} \][/tex]
4. This gives us two potential solutions:
[tex]\[ w = \frac{-3 + 21}{4} = \frac{18}{4} = 4.5 \][/tex]
And
[tex]\[ w = \frac{-3 - 21}{4} = \frac{-24}{4} = -6 \][/tex]
Since a width cannot be negative, we discard [tex]\( w = -6 \)[/tex] and accept [tex]\( w = 4.5 \)[/tex] meters.
Now, we substitute [tex]\( w \)[/tex] back into the expression for the length:
[tex]\[ \text{Length} = 2w + 3 \][/tex]
[tex]\[ \text{Length} = 2(4.5) + 3 = 9 + 3 = 12 \, \text{meters} \][/tex]
Thus, the dimensions of the rectangle are:
- Width: [tex]\( 4.5 \)[/tex] meters
- Length: [tex]\( 12 \)[/tex] meters
So, the results are:
Length: [tex]\( 12 \)[/tex] meters
Width: [tex]\( 4.5 \)[/tex] meters
