Westonci.ca is the ultimate Q&A platform, offering detailed and reliable answers from a knowledgeable community. Get immediate and reliable answers to your questions from a community of experienced professionals on our platform. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
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
We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.
