Write a complete solution to each problem.
If the length of a rectangle is 3 feet longer than the width and
the diagonal I s 15 feet, then what are the length and width?
I do not understand how to solve this problem. Can someone help me?
Thanks.
12x9
Use the Pythagorean theorem.
Hypotenuse^2 = sum of squares of other two sides
Let x = length, then width = x - 3
15^2 = x^2 + (x-3)^2
Does this help?
To solve this problem, we need to use the Pythagorean theorem and the given information.
1. Let's start by assigning variables to the unknown values. Let:
- W = width of the rectangle
- L = length of the rectangle
2. Based on the given information, we know that the length is 3 feet longer than the width. Therefore, we can write an equation:
L = W + 3
3. We also know that the diagonal of the rectangle is 15 feet. According to the Pythagorean theorem, the square of the length of the diagonal is equal to the sum of the squares of the length and width. So, we can write another equation:
W^2 + L^2 = diagonal^2
W^2 + (W + 3)^2 = 15^2
4. Simplify and solve the equation:
W^2 + (W^2 + 6W + 9) = 225
2W^2 + 6W + 9 - 225 = 0
2W^2 + 6W - 216 = 0
Divide every term by 2: W^2 + 3W - 108 = 0
To solve this quadratic equation, we can either factor it or use the quadratic formula.
5. Factoring the quadratic equation:
(W - 9)(W + 12) = 0
Setting each factor equal to zero gives us two possible solutions:
W - 9 = 0 -> W = 9
W + 12 = 0 -> W = -12
Since a negative width doesn't make sense in this context, we discard the negative solution.
6. Now that we have a value for W, we can substitute it back into the equation for L:
L = W + 3 = 9 + 3 = 12
Therefore, the width of the rectangle is 9 feet and the length is 12 feet.