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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.
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