IDNLearn.com provides a seamless experience for finding and sharing answers. Our community is here to provide the comprehensive and accurate answers you need to make informed decisions.
Sagot :
Sure, let's solve this step-by-step.
1. Understand the problem:
- We have a rectangular park with an area of [tex]\(180 \, \text{m}^2\)[/tex].
- The park is fenced on three sides with a total fencing length of 39 meters, leaving one of the longer sides unfenced.
2. Define the variables:
- Let [tex]\(x\)[/tex] represent the width of the rectangle (in meters).
- Let [tex]\(y\)[/tex] represent the length of the rectangle (in meters).
3. Given relationships:
- The area of the rectangle is given by the product of its length and width: [tex]\(x \times y = 180\)[/tex].
- The total length of the fence used is for the two widths and one length, so: [tex]\(2x + y = 39\)[/tex].
4. Express the length [tex]\(y\)[/tex] in terms of width [tex]\(x\)[/tex] using the area equation:
- From [tex]\(x \times y = 180\)[/tex], we can express [tex]\(y\)[/tex] as:
[tex]\[ y = \frac{180}{x} \][/tex]
5. Substitute [tex]\(y\)[/tex] in the fencing equation:
- Replace [tex]\(y\)[/tex] in the fencing equation [tex]\(2x + y = 39\)[/tex]:
[tex]\[ 2x + \frac{180}{x} = 39 \][/tex]
6. Solve for [tex]\(x\)[/tex]:
- To solve this equation, we can multiply through by [tex]\(x\)[/tex] to clear the fraction:
[tex]\[ 2x^2 + 180 = 39x \][/tex]
- Rearrange the equation into standard quadratic form:
[tex]\[ 2x^2 - 39x + 180 = 0 \][/tex]
7. Apply the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 2\)[/tex], [tex]\(b = -39\)[/tex], and [tex]\(c = 180\)[/tex]:
8. Calculate the discriminant:
- [tex]\( \Delta = b^2 - 4ac \)[/tex]
- [tex]\( \Delta = (-39)^2 - 4(2)(180) \)[/tex]
- [tex]\( \Delta = 1521 - 1440 = 81 \)[/tex]
9. Calculate [tex]\(x\)[/tex]:
[tex]\[ x = \frac{39 \pm \sqrt{81}}{4} \][/tex]
[tex]\[ x = \frac{39 \pm 9}{4} \][/tex]
This gives two potential solutions:
[tex]\[ x = \frac{48}{4} = 12 \][/tex]
[tex]\[ x = \frac{30}{4} = 7.5 \][/tex]
10. Determine the corresponding [tex]\(y\)[/tex] values for each [tex]\(x\)[/tex]:
For [tex]\(x = 12\)[/tex]:
[tex]\[ y = \frac{180}{12} = 15 \][/tex]
For [tex]\(x = 7.5\)[/tex]:
[tex]\[ y = \frac{180}{7.5} = 24 \][/tex]
Since one of the sides (length) has to be unfenced, the feasible solution that fits the condition (leaving one of the longer sides unfenced) is:
[tex]\[ x = 7.5 \, \text{m} \quad \text{(width)}, \][/tex]
[tex]\[ y = 24 \, \text{m} \quad \text{(length)}. \][/tex]
Thus, the dimensions of the park are 7.5 meters in width and 24 meters in length.
1. Understand the problem:
- We have a rectangular park with an area of [tex]\(180 \, \text{m}^2\)[/tex].
- The park is fenced on three sides with a total fencing length of 39 meters, leaving one of the longer sides unfenced.
2. Define the variables:
- Let [tex]\(x\)[/tex] represent the width of the rectangle (in meters).
- Let [tex]\(y\)[/tex] represent the length of the rectangle (in meters).
3. Given relationships:
- The area of the rectangle is given by the product of its length and width: [tex]\(x \times y = 180\)[/tex].
- The total length of the fence used is for the two widths and one length, so: [tex]\(2x + y = 39\)[/tex].
4. Express the length [tex]\(y\)[/tex] in terms of width [tex]\(x\)[/tex] using the area equation:
- From [tex]\(x \times y = 180\)[/tex], we can express [tex]\(y\)[/tex] as:
[tex]\[ y = \frac{180}{x} \][/tex]
5. Substitute [tex]\(y\)[/tex] in the fencing equation:
- Replace [tex]\(y\)[/tex] in the fencing equation [tex]\(2x + y = 39\)[/tex]:
[tex]\[ 2x + \frac{180}{x} = 39 \][/tex]
6. Solve for [tex]\(x\)[/tex]:
- To solve this equation, we can multiply through by [tex]\(x\)[/tex] to clear the fraction:
[tex]\[ 2x^2 + 180 = 39x \][/tex]
- Rearrange the equation into standard quadratic form:
[tex]\[ 2x^2 - 39x + 180 = 0 \][/tex]
7. Apply the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 2\)[/tex], [tex]\(b = -39\)[/tex], and [tex]\(c = 180\)[/tex]:
8. Calculate the discriminant:
- [tex]\( \Delta = b^2 - 4ac \)[/tex]
- [tex]\( \Delta = (-39)^2 - 4(2)(180) \)[/tex]
- [tex]\( \Delta = 1521 - 1440 = 81 \)[/tex]
9. Calculate [tex]\(x\)[/tex]:
[tex]\[ x = \frac{39 \pm \sqrt{81}}{4} \][/tex]
[tex]\[ x = \frac{39 \pm 9}{4} \][/tex]
This gives two potential solutions:
[tex]\[ x = \frac{48}{4} = 12 \][/tex]
[tex]\[ x = \frac{30}{4} = 7.5 \][/tex]
10. Determine the corresponding [tex]\(y\)[/tex] values for each [tex]\(x\)[/tex]:
For [tex]\(x = 12\)[/tex]:
[tex]\[ y = \frac{180}{12} = 15 \][/tex]
For [tex]\(x = 7.5\)[/tex]:
[tex]\[ y = \frac{180}{7.5} = 24 \][/tex]
Since one of the sides (length) has to be unfenced, the feasible solution that fits the condition (leaving one of the longer sides unfenced) is:
[tex]\[ x = 7.5 \, \text{m} \quad \text{(width)}, \][/tex]
[tex]\[ y = 24 \, \text{m} \quad \text{(length)}. \][/tex]
Thus, the dimensions of the park are 7.5 meters in width and 24 meters in length.
We appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. Find clear and concise answers at IDNLearn.com. Thanks for stopping by, and come back for more dependable solutions.
