A rectangular parcel of land has an area of 2,000 ft2. A diagonal between opposite corners is measured to be 10 ft longer than one side of the parcel. What are the dimensions of the land, correct to the nearest foot?
To solve this problem, let's start by assuming that the length of the rectangular parcel is L ft and the width is W ft.
We are given that the area of the parcel is 2,000 ft², so we can write the equation:
L * W = 2000
We are also given that the diagonal between opposite corners is measured to be 10 ft longer than one side of the parcel. Let's assume the width is W ft, so the length of the diagonal can be represented as W + 10 ft.
Now, we can use the Pythagorean theorem to find the relationship between the length, width, and diagonal. According to the Pythagorean theorem, the square of the diagonal is equal to the sum of the squares of the other two sides.
So, we have the equation:
L² + W² = (W + 10)²
Expanding this equation:
L² + W² = W² + 20W + 100
Simplifying the equation:
L² = 20W + 100
Now, let's substitute the value of L from the first equation into the second equation:
(W * W) = 20W + 100
This is a quadratic equation. Rearranging it:
W² - 20W - 100 = 0
Now we can solve this quadratic equation to find the width (W) of the rectangle. We can factorize it or use the quadratic formula.
By factoring, we can rewrite the quadratic equation as:
(W - 10)(W + 10) - 100 = 0
Simplifying:
(W - 10)(W + 10) = 100
Now, we have two possible solutions for W:
1) W - 10 = 10 and W + 10 = 100,
Solving for W, we get W = 20
2) W - 10 = 100 and W + 10 = 10
Solving for W, we get W = 110
Since the dimensions of the land cannot be negative, we can discard the second solution. Therefore, we have W = 20 ft.
Now, substitute the value of W into the first equation to find L:
L = 2000 ÷ 20 = 100 ft
Therefore, the dimensions of the rectangular parcel of land are approximately 100 ft by 20 ft.