Westonci.ca is your go-to source for answers, with a community ready to provide accurate and timely information. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
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
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.