A rectangular parcel of land has an area of 4,000 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?
x^2 + y^2 = diagonal^2 = d^2
x y = 4000 so y = 4000/x
d = x+10 so d^2 = x^2 + 20 x + 100
x^2 + 16*10^6/x^2 = x^2 + 20 x + 100
16*10^6 = 20 x^3 + 100 x^2
5 x^3 + 25 x^2 - 4*10^6 = 0
I do not see an easy way to solve that. Use iteration, x and 5x^3+25 x^2
x = 10^2 --> 5,250,000 too big
x = 80 --> 2,720,000 too small
x = 90 --> 3,847,500 getting close
x = 92 --> 4,333,568 very close
x = 91 --> 3,974,880 between 91 and 92
x = 91.5-> 4,039,610 between 91 and 91.5
so 91 to nearest foot
91^2 + y^2 = (91+10))^2
8281 +y^2 = 10201
y = 43.8 or 44
so
91 by 44
let one side be x
then the diagonal is x+10
let the third side by y
x^2 + y^2 = (x+10)^2
x^2 + y^2 = x^2 + 20x + 100
y^2 = 20x + 100
y = √(20x+100)
so the area of the triangle is half the rectangle area
(1/2)xy =(1/2)(4000)
xy = 4000
x√(20x+100) = 4000
x^2(20x+100) = 16000000
20x^3 + 100x^2 - 16000000 = 0
x^3 + 5x^2 - 800000 = 0
I must admit that I used Wolfram to solve this
http://www.wolframalpha.com/input/?i=x%5E3+%2B+5x%5E2+-+800000+%3D+0
and I got x = 91.1947
then y = 43.8622
and the diagonal would be 101.1947
check:
area of rectangle = xy = (91.1947)(43.8622) = 4000.00017 (close enough)
I will let you check if Pythagoras also works out, it does.
I don't know what method you would have to solve that cubic, since it does not factor, and therefore does not have rational roots.
Do you know Newton's Method ?
Well, well, well, it seems we have a rectangular puzzle on our hands! Let's solve this riddle, shall we?
Let's assume the length of the rectangle is L and the width is W. We are given that the area of the parcel is 4,000 ft², which means L * W = 4,000.
Now, we are also told that the length of the diagonal (let's call it D) is 10 ft longer than one side of the parcel, meaning D = L + 10.
Alright, let's put our clown hats on and dive into some clown math!
Since we have a right-angled triangle formed by the diagonal, length, and width, we can use the Pythagorean theorem to solve this jibber-jabber.
According to Pythagoras, the sum of the squares of the sides of a right-angled triangle is equal to the square of the hypotenuse. So, L² + W² = D².
Substituting in the values we know, we get L² + W² = (L + 10)².
Expanding that equation, we have L² + W² = L² + 20L + 100.
Now subtracting L² from both sides, we get W² = 20L + 100.
We also know that L * W = 4,000, so we can rewrite that as L * (20L + 100) = 4,000.
Expanding that equation, we get 20L² + 100L - 4,000 = 0.
Now, let's put our brains to the test and solve this quadratic equation by factoring or using the quadratic formula.
After some calculations, we find that the length of the rectangle, L, is approximately 31 feet.
Using the area formula, we can find the width, W, as 4,000 divided by the length, which is roughly 129 feet.
So, to wrap up this circus act, the dimensions of the land are approximately 31 feet by 129 feet. Ta-da!
To find the dimensions of the land, we need to solve this problem step-by-step.
Step 1: Determine the length and width of the parcel.
Let's assume the length of the parcel is L ft and the width is W ft.
Step 2: Find the equation that represents the area of the parcel.
Since the area of a rectangle is given by A = length × width, we have A = L × W.
Given that the area is 4,000 ft^2, we have 4,000 = L × W.
Step 3: Find the relationship between the diagonal and the sides of the parcel.
According to the problem, the diagonal between opposite corners is measured to be 10 ft longer than one side of the parcel. This relationship can be expressed as:
diagonal = L + 10 (Equation 1)
Step 4: Find the relationship between the diagonal and the sides using the Pythagorean theorem.
According to the Pythagorean theorem, in a right-angled triangle, the square of the hypotenuse (diagonal) is equal to the sum of the squares of the other two sides (length and width). We can write this as:
diagonal^2 = L^2 + W^2 (Equation 2)
Step 5: Solve the system of equations formed by Equations 1 and 2.
Substitute the value of diagonal from Equation 1 into Equation 2:
(L + 10)^2 = L^2 + W^2
Expand the equation:
L^2 + 20L + 100 = L^2 + W^2
Simplify the equation by canceling out the common terms:
20L + 100 = W^2
Step 6: Substitute the area equation (4,000 = L × W) into the equation obtained in Step 5.
Replace W^2 in the equation with (4,000/L):
20L + 100 = (4,000/L)
Step 7: Solve the equation obtained in Step 6 to find the value of L.
Multiply through by L to eliminate the fraction:
20L^2 + 100L - 4,000 = 0
Solve this quadratic equation for L using the quadratic formula. The quadratic formula is given by:
L = (-b ± √(b^2 - 4ac)) / 2a
where a = 20, b = 100, and c = -4,000.
Solving this equation gives two possible solutions for L: L ≈ 33.57 ft or L ≈ -6.04 ft.
Since the length of the land cannot be negative, we take the positive value, L ≈ 33.57 ft.
Step 8: Substitute the value of L into the area equation (4,000 = L × W) to find the value of W.
W = 4,000 / L = 4,000 / 33.57 ≈ 119.08 ft
Thus, the dimensions of the land are approximately:
Length (L) ≈ 33.57 ft
Width (W) ≈ 119.08 ft
To find the dimensions of the land, we can use the information given about the area and the diagonal. Let's proceed step by step:
1. Let's assume the length of the rectangular parcel is "L" and the width is "W".
2. Since the area of the parcel is given as 4,000 ft², we have the equation L x W = 4,000.
3. We are also given that the diagonal (let's call it "D") is measured to be 10 ft longer than one side of the parcel. So, we have D = L + 10.
4. We can use the Pythagorean theorem to relate the lengths of the sides and the diagonal of a rectangle. The theorem states that the square of the length of the diagonal is equal to the sum of the squares of the lengths of the sides. In this case, we have D² = L² + W².
5. Substituting D = L + 10 into the Pythagorean equation, we get (L + 10)² = L² + W².
6. Expanding the equation, we have L² + 20L + 100 = L² + W².
7. Rearranging the equation, we get W² = 20L + 100.
8. Now, we can substitute the value of W² from equation 7 into the area equation (step 2): L x (20L + 100) = 4,000.
9. Expanding the equation, we have 20L² + 100L - 4,000 = 0.
10. Dividing the equation by 20 to simplify, we have L² + 5L - 200 = 0.
11. This is a quadratic equation that can be factored as (L + 20)(L - 10) = 0.
12. Therefore, L + 20 = 0 or L - 10 = 0.
13. If L + 20 = 0, then L = -20, which is not a valid dimension.
14. If L - 10 = 0, then L = 10.
15. Substituting L = 10 into equation 2, we have 10 x W = 4,000.
16. Therefore, W = 4,000 / 10 = 400 ft.
17. Finally, the dimensions of the land are L = 10 ft and W = 400 ft.
So, the correct dimensions of the rectangular land are 10 ft by 400 ft.