Suppose you want to build a rectangular sandbox where the width is 5 feet less than the length and the diagonal is 5 feet longer than the length. What are the dimensions of the sandbox?
If the length is x, then you must have
(x-5)^2 + x^2 = (x+5)^2
You can work through the math, but just think of the simple
3-4-5 right triangle. Scale it up by a factor of 5, and you have
15-20-25
Length = L.
Width = L-5 ft.
Diag. = L+5 ft.
L^2 + (L-5)^2 = (L+5)^2.
L^2 + L^2 - 10L +25 = L^2 + 10L + 25,
L^2 - 20L = 0,
Divide both sides by L:
L - 20 = 0,
L = 20 ft.
Width = L-5 = 15 ft.
To find the dimensions of the rectangular sandbox, we can set up a system of equations based on the given information. Let's call the length of the sandbox "L" and the width "W."
1. From the given information, we know that the width is 5 feet less than the length. This can be expressed as:
W = L - 5
2. We are also given that the diagonal is 5 feet longer than the length. Using the Pythagorean theorem, we can find the relationship between the length, width, and diagonal:
L^2 + W^2 = (L + 5)^2
Now, let's solve the system of equations to find the values of L and W:
Substituting the value of W from the first equation into the second equation:
L^2 + (L - 5)^2 = (L + 5)^2
Expanding the equation:
L^2 + (L^2 - 10L + 25) = L^2 + 10L + 25
Combining like terms:
2L^2 - 10L + 25 = L^2 + 10L + 25
Subtracting L^2 and 25 from both sides:
L^2 - 20L = 0
Factoring out L:
L(L - 20) = 0
Setting each factor equal to zero and solving for L:
L = 0 or L - 20 = 0
Since the length cannot be zero, we have:
L - 20 = 0
L = 20
Now that we have the value of L, we can substitute it back into the first equation to find W:
W = L - 5
W = 20 - 5
W = 15
Therefore, the dimensions of the sandbox are length = 20 feet and width = 15 feet.
To solve this problem, we can use the Pythagorean theorem to relate the length, width, and diagonal of the sandbox. The Pythagorean theorem states that in a right triangle, the square of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides.
Let's assign variables to represent the length and width of the sandbox. We can call the length "L" and the width "W".
From the problem statement, we know that the width is 5 feet less than the length. So we have the equation:
W = L - 5
We also know that the diagonal is 5 feet longer than the length. This can be expressed as:
√(L^2 + W^2) = L + 5
Now we have a system of two equations with two variables. By substituting the value of W from the first equation into the second equation, we can solve for L.
Substituting W = L - 5 into the second equation:
√(L^2 + (L - 5)^2) = L + 5
Simplifying this equation, we get:
√(L^2 + L^2 - 10L + 25) = L + 5
√(2L^2 - 10L + 25) = L + 5
Squaring both sides of the equation to eliminate the square root:
2L^2 - 10L + 25 = (L + 5)^2
2L^2 - 10L + 25 = L^2 + 10L + 25
2L^2 - 10L = L^2 + 10L
L^2 - 20L = 0
Factorizing both sides:
L(L - 20) = 0
So we have two possible solutions:
L = 0 (rejecting this since it does not make sense in this context)
L - 20 = 0
Solving for L:
L = 20
Now, we can substitute the value of L back into the first equation to solve for W:
W = L - 5
W = 20 - 5
W = 15
Therefore, the dimensions of the sandbox are length = 20 feet and width = 15 feet.