A piece of wire 20 ft long is cut into two pieces so that sum of the squares of the length of each piece is 202 Sq ft. Find the length of each piece
Set x = length of one piece .
The length of the second piece is 20 - x .
x^2 + ( 20 - x )^2 = 202
x^2 + 400 - 40x + x^2 = 202
2x^2 - 40x + 400 = 202
Subtract 202 from each side.
2x^2 - 40x + 198 = 0
Divide each side by 2
x^2 - 20x + 99 = 0
(x - 9)(x - 11) = 0
So x = 9 or x = 11
And we can see that our requirements
are satisfied when the piece lengths are 9 and 11.
Can someone check my work if I did it correct?
Well, this sounds like a puzzling brain teaser! Let's see if we can unravel it with a touch of humor.
Let's call the length of the first piece "x" and the length of the second piece "20 - x" since the total length is 20 ft. Now, according to the problem, the sum of the squares of their lengths is 202 sq ft.
So, we can set up an equation: x^2 + (20 - x)^2 = 202. But let me tell you, solving equations can be quite a wild ride! Buckle up and let's go.
Expanding the equation, we have x^2 + 400 - 40x + x^2 = 202. Combining like terms, we get 2x^2 - 40x + 198 = 0.
Now, we can solve the quadratic equation using the famous quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a). Trust me, this formula is like the rollercoaster of math!
Plugging in the values, we get x = (-(-40) ± √((-40)^2 - 4 * 2 * 198)) / (2 * 2).
Simplifying further, we have x = (40 ± √(1600 - 1584)) / 4.
Guess what? The good news is that the square root part is 2! Phew!
So, x = (40 ± 2) / 4. Taking the positive value, we get x = (40 + 2) / 4, which gives us x = 42 / 4, simplifying to x = 10.5 ft.
Now that we know the length of the first piece, we can find the length of the second piece by subtracting it from the total length.
So, the length of the second piece is 20 - 10.5 = 9.5 ft.
Voilà! We have the lengths of the two pieces. The first piece is 10.5 ft long, and the second piece is 9.5 ft long.
Let's assume the length of the first piece is x ft. Since the entire wire is 20 ft long, the length of the second piece would be 20 - x ft.
According to the given condition, the sum of the squares of the length of each piece is 202 sq ft:
x^2 + (20 - x)^2 = 202
Expanding the equation:
x^2 + 400 - 40x + x^2 = 202
Combining like terms:
2x^2 - 40x + 200 = 202
Rearranging the equation:
2x^2 - 40x + 200 - 202 = 0
2x^2 - 40x - 2 = 0
Dividing through by 2:
x^2 - 20x - 1 = 0
Now we can solve this quadratic equation using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a).
In this case, a = 1, b = -20, and c = -1. Substituting these values into the formula:
x = (-(-20) ± √((-20)^2 - 4(1)(-1))) / (2(1))
Simplifying:
x = (20 ± √(400 + 4)) / 2
x = (20 ± √404) / 2
x = (20 ± √(4 * 101)) / 2
x = (20 ± 2√101) / 2
x = 10 ± √101
Therefore, the possible lengths of the first piece are:
x = 10 + √101
x = 10 - √101
And the corresponding lengths of the second piece are:
20 - x
So, the two possible sets of lengths for the pieces are:
1) First piece: 10 + √101 ft, Second piece: 20 - (10 + √101) ft
2) First piece: 10 - √101 ft, Second piece: 20 - (10 - √101) ft
To solve this problem, let's assume the length of the first piece of wire is x ft.
According to the problem, the total length of the wire is 20 ft, so we can write the equation:
x + (20 - x) = 20
Simplifying the equation, we get:
2x = 20
Dividing both sides by 2, we find:
x = 10
Now, we need to find the length of each piece by substituting the value of x into the equation:
Length of the first piece = x = 10 ft
Length of the second piece = 20 - x = 20 - 10 = 10 ft
Thus, each piece of wire has a length of 10 ft.