Problem 48
A garden area is 30ft long and 20ft wide. A path of uniform width is set around the edge. If the remaining garden area is 400ft ^2, what is the width of the path?
My answer: 2 radical (2)i
your answer is correct.. you are very good at this stuff. your gonna be an expert..
Thank you for your kind words! I'm here to assist you further, so let's work through the problem together.
To find the width of the path, we'll need to subtract the area of the remaining garden from the total area of the garden with the path. Let's start by finding the total area of the garden with the path.
The total length of the garden, including the path, will be the original length plus twice the width of the path. In this case, it will be 30ft + 2w, where w is the width of the path.
Similarly, the total width of the garden will be the original width plus twice the width of the path. In this case, it will be 20ft + 2w.
The area of a rectangle is given by the product of its length and width. So, the total area of the garden with the path is:
(30ft + 2w) * (20ft + 2w)
Now, we are given that the remaining garden area is 400ft^2. So, we can set up the following equation:
(30ft + 2w) * (20ft + 2w) - 400ft^2 = 0
Expanding the equation and rearranging it to the standard quadratic form, we get:
4w^2 + 100w - 200 = 0
Now we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. In this case, factoring might not be easy, so let's use the quadratic formula:
w = (-100 ± √(100^2 - 4 * 4 * -200)) / (2 * 4)
Simplifying this expression gives:
w = (-100 ± √(10000 + 3200)) / 8
w = (-100 ± √(13200)) / 8
Now, we need to simplify the expression under the square root:
√(13200) = √(400 * 33) = 20√33
So the equation further simplifies to:
w = (-100 ± 20√33) / 8
Now, we can simplify this expression by dividing both the numerator and denominator by 4:
w = (-25 ± 5√33) / 2
And this is the final answer for the width of the path.
Therefore, your answer of 2√2i is not correct for this problem. The correct answer for the width of the path is (-25 ± 5√33) / 2.