SOLUTION: To greet the 2020 SHS graduates, a tarpaulin is to be set along the national highway. If the area of the tarp is to be 35/4 m^2 and its perimeter 27/2 meters, what should be the di

Question 1186463: To greet the 2020 SHS graduates, a tarpaulin is to be set along the national highway. If the area of the tarp is to be 35/4 m^2 and its perimeter 27/2 meters, what should be the dimensions of the tarpaulin

Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Here's how to find the dimensions of the tarpaulin:
**1. Set up the equations:**
Let 'w' be the width and 'h' be the height of the tarpaulin. We are given the area and the perimeter:
* Area: w * h = 35/4
* Perimeter: 2w + 2h = 27/2
**2. Solve for one variable in terms of the other:**
From the area equation, we can express h in terms of w:
h = (35/4) / w
**3. Substitute into the perimeter equation:**
2w + 2 * ((35/4) / w) = 27/2
**4. Simplify and solve for w:**
2w + (35/2w) = 27/2
Multiply the entire equation by 2w to get rid of the fraction:
4w² + 35 = 27w
4w² - 27w + 35 = 0
**5. Solve the quadratic equation:**
We can use the quadratic formula to solve for w. The formula is:
w = (-b ± sqrt(b² - 4ac)) / 2a
In our equation, a = 4, b = -27, and c = 35.
w = (27 ± sqrt((-27)² - 4 * 4 * 35)) / (2 * 4)
w = (27 ± sqrt(729 - 560)) / 8
w = (27 ± sqrt(169)) / 8
w = (27 ± 13) / 8
This gives us two possible solutions for w:
w = (27 + 13) / 8 = 40 / 8 = 5
w = (27 - 13) / 8 = 14 / 8 = 7/4 = 1.75
**6. Find the corresponding values for h:**
If w = 5, then h = (35/4) / 5 = 35/20 = 7/4 = 1.75
If w = 1.75, then h = (35/4) / 1.75 = 5
**7. State the dimensions:**
The dimensions of the tarpaulin are 5 meters by 1.75 meters. It doesn't matter which dimension is width or height.

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