A rectangular chicken yard was built against an existing barn wall (as shown in the picture). 30 m of fencing was used to enclose the yard.
a) Find the dimensions of the yard its area is 108 m!
How to solve this problem..?
xy = 108
x+2y = 30
x + 2(108/x) = 30
x^2 - 30x + 216 = 0
(x-12)(x-18) = 0
The yard is 12x9
To solve this problem, we can use the information provided to write equations and solve for the dimensions of the yard. Here’s how you can approach it step-by-step:
1. Let's assume the width of the yard is "w" meters and the length of the yard is "l" meters.
2. Since the yard is rectangular and is built against the barn wall, we know that the width will be larger than the length.
3. The perimeter of the yard is given as 30 meters, which can be expressed as: 2w + l + w = 30.
- The first term, 2w, represents the two widths on either side.
- The second term, l, represents the length.
- The third term, w, represents the width against the barn wall.
4. Simplifying the equation: 2w + l + w = 30 becomes 3w + l = 30.
Now, we can proceed to find the dimensions of the yard its area is 108 m².
5. The formula for the area of a rectangle is given by: Area = length × width.
6. In this case, the area is given as 108 m², so we have the equation l × w = 108.
7. We have two equations now: 3w + l = 30 and l × w = 108.
Now we can solve these equations simultaneously to find the values of l and w.
8. Rearrange the first equation to express l in terms of w: l = 30 - 3w.
9. Substitute this value of l in the second equation: w × (30 - 3w) = 108.
10. Distribute the multiplication: 30w - 3w² = 108.
11. Rearrange the equation to form a quadratic equation: 3w² - 30w + 108 = 0.
12. Divide the entire equation by 3 to simplify it: w² - 10w + 36 = 0.
Now we can solve this quadratic equation using factoring, completing the square, or the quadratic formula.
13. Factor the quadratic equation: (w - 4)(w - 9) = 0.
14. Set each factor equal to zero to solve for w: w - 4 = 0 or w - 9 = 0.
15. Solve for w: w = 4 or w = 9.
Based on the given information, the width cannot be greater than the total perimeter, so the width is 4 meters.
Now we can find the length:
16. Substitute the value of w in one of the previous equations, e.g., 3w + l = 30.
17. Plug in the value w = 4: 3(4) + l = 30.
18. Solve for l: 12 + l = 30.
19. Subtract 12 from both sides: l = 18.
Therefore, the dimensions of the yard are 4 meters by 18 meters.
To solve this problem, we need to find the dimensions of the rectangular chicken yard, given that its area is 108 square meters and 30 meters of fencing were used to enclose it.
Let's assume the length of the yard is L meters and the width is W meters.
1. Draw a diagram: Sketch a rectangle representing the chicken yard against the barn wall. Label the length as L and the width as W.
2. Write down the formulas: The perimeter of a rectangle is given by the formula P = 2L + 2W, and the area is given by the formula A = L * W.
3. Use the given information: We know that the total length of the fence used is 30 meters, so the perimeter of the rectangle is 30 meters. Thus, we have the equation 2L + 2W = 30.
4. Use the area formula: We are also given that the area of the yard is 108 square meters, so we have the equation L * W = 108.
5. Solve the equations: We have a system of two equations, 2L + 2W = 30 and L * W = 108. We can solve this system using substitution or elimination.
Using elimination:
- Multiply the first equation by 2 to eliminate the L term: 4L + 4W = 60.
- Subtract the second equation from the modified first equation: (4L + 4W) - (2L + 2W) = 60 - 30.
- Simplify: 2L + 2W = 30, so 2L + 2W - 2L - 2W = 30 - 108, which simplifies to 0 = -78.
- Since this is not a valid equation, there must have been a mistake in the problem setup. Please check the given information again to see if there was an error in the values provided.
If the given values are correct, it may not be possible to create a rectangular chicken yard with an area of 108 square meters and a fence length of 30 meters.