A rectangular piece of cardboard has length that is 16 inches more than the width. The area of the cardboard is 297 sq. in. Find the measures of its width and length.
Enter width in., length in.
L = W + 16
W(W+16) = 297
Solve for W, then L.
the area of the rectangle is 22 square feet, if its width is 3 1/3 feet, find its length
To solve this problem, we can set up an equation based on the given information.
Let's say the width of the cardboard is 'w' inches.
According to the problem, the length of the cardboard is 16 inches more than the width. So, the length would be 'w + 16' inches.
The area of a rectangle is given by the formula: Area = length × width.
So we have the equation:
w × (w + 16) = 297
To solve this equation, we can multiply out the terms:
w^2 + 16w = 297
Now, we need to rearrange this equation into standard form, where one side is equal to zero:
w^2 + 16w - 297 = 0
To solve this quadratic equation, we can factor it or use the quadratic formula. Factoring might not be obvious in this case, so let's use the quadratic formula:
w = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a = 1, b = 16, and c = -297. Plugging these values into the quadratic formula, we get:
w = (-16 ± √(16^2 - 4(1)(-297))) / (2(1))
Simplifying:
w = (-16 ± √(256 + 1188)) / 2
w = (-16 ± √1444) / 2
w = (-16 ± 38) / 2
Now, we have two possible values for the width, which are:
w1 = (-16 + 38) / 2 = 11 inches
w2 = (-16 - 38) / 2 = -27 inches (not a valid solution since the width cannot be negative)
So, the width of the cardboard is 11 inches.
To find the length, we can substitute the value of the width into the equation we set up earlier:
Length = Width + 16
Length = 11 + 16
Length = 27 inches
Therefore, the measures of the width and length of the cardboard are 11 inches and 27 inches, respectively.