PROBLEM: As an engineer you are tasked with designing your clients dream house. Your client wants to know the size of their dream house, given the area should be 52 square meters. additonally they have specified the length of the house should be 4 meters more than the width. what should be the dimension of the dream house?
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Answer:
To find the dimensions of your client's dream house with an area of 52 square meters, where the length is 4 meters more than the width, you can set up an equation.
Let's use the width as "W" meters. According to the problem, the length is 4 meters more than the width, so the length is "W + 4" meters.
The formula for the area of a rectangle is:
Area = Length x Width
So, in this case:
52 square meters = (W + 4) meters x W meters
Now, you can solve for W:
W² + 4W - 52 = 0
To solve this quadratic equation, you can use the quadratic formula:
W = (-b ± √(b² - 4ac)) / (2a)
In this case, a = 1, b = 4, and c = -52. Plugging these values into the quadratic formula:
W = (-4 ± √(4² - 4(1)(-52))) / (2(1))
W = (-4 ± √(16 + 208)) / 2
W = (-4 ± √224) / 2
Now, you can simplify the square root:
W = (-4 ± 4√14) / 2
W = -2 ± 2√14
Now you have two possible values for W:
1. W = -2 + 2√14
2. W = -2 - 2√14
Since the width cannot be negative in this context, you can discard the negative solution:
W = -2 + 2√14
Now, you have the width, and you can find the length:
Length = W + 4
Length = (-2 + 2√14) + 4
So, the width is approximately -2 + 2√14 meters, and the length is approximately 2√14 + 2 meters. These are the dimensions of your client's dream house with an area of 52 square meters.