The length of a rectangle is 2 inches longer than the width. The area of the rectangle is 56 in^2. Find the dimensions.
Also: The length of a rectangle is 3 ft more than the width. The area of the rectangle is 28 ft^2. Find the dimensions.
w(w+2) = 56
w^2+2w-56 = 0
w = √57 - 1
To find the dimensions of the rectangle, we can set up an equation.
Let's say the width of the rectangle is x inches.
According to the problem, the length of the rectangle is 2 inches longer than the width, so the length would be x + 2 inches.
The area of a rectangle is given by the formula: area = length * width.
Given that the area of the rectangle is 56 square inches, we can set up the equation:
56 = (x + 2) * x
56 = x^2 + 2x
x^2 + 2x - 56 = 0
Now we can solve this quadratic equation for x.
Factoring or using the quadratic formula, we find that x = 6 and x = -8.
Since dimensions cannot be negative, the width of the rectangle is 6 inches.
The length would be 6 + 2 = 8 inches.
So, the dimensions of the rectangle are 6 inches by 8 inches.
Now, let's solve the second problem.
Let's say the width of the rectangle is y feet.
According to the problem, the length of the rectangle is 3 feet more than the width, so the length would be y + 3 feet.
The area of a rectangle is given by the formula: area = length * width.
Given that the area of the rectangle is 28 square feet, we can set up the equation:
28 = (y + 3) * y
28 = y^2 + 3y
y^2 + 3y - 28 = 0
Now we can solve this quadratic equation for y.
Factoring or using the quadratic formula, we find that y = 4 and y = -7.
Since dimensions cannot be negative, the width of the rectangle is 4 feet.
The length would be 4 + 3 = 7 feet.
So, the dimensions of the rectangle are 4 feet by 7 feet.
To find the dimensions of the rectangle, we can use the given information about the length, width, and area of the rectangle.
For the first question, let's denote the width of the rectangle as "w" inches. According to the problem, the length of the rectangle is 2 inches longer than the width, so we can express the length as "w + 2" inches.
The area of a rectangle is calculated by multiplying the length by the width. In this case, we know that the area is 56 in^2. Therefore, we can set up the equation:
Area = Length * Width
56 = (w + 2) * w
Expanding the equation:
56 = w^2 + 2w
Rearranging the equation to the quadratic form:
w^2 + 2w - 56 = 0
Now we can solve this quadratic equation for "w" by factoring or using the quadratic formula.
Factoring:
(w + 8)(w - 7) = 0
Setting each bracket equal to zero:
w + 8 = 0 or w - 7 = 0
Solving for "w":
w = -8 or w = 7
Since the width of the rectangle cannot be negative, we discard w = -8. Therefore, the width of the rectangle is 7 inches.
Using this value, we can find the length by adding 2 inches:
Length = Width + 2 = 7 + 2 = 9 inches
So, the dimensions of the rectangle are 7 inches by 9 inches.
For the second question, we can follow a similar approach.
Let's denote the width of the rectangle as "w" feet. According to the problem, the length of the rectangle is 3 feet more than the width, so we can express the length as "w + 3" feet.
The area of a rectangle is calculated by multiplying the length by the width. In this case, we know that the area is 28 ft^2. Therefore, we can set up the equation:
Area = Length * Width
28 = (w + 3) * w
Expanding the equation:
28 = w^2 + 3w
Rearranging the equation to the quadratic form:
w^2 + 3w - 28 = 0
Again, we can solve this quadratic equation for "w" by factoring or using the quadratic formula.
Factoring:
(w + 7)(w - 4) = 0
Setting each bracket equal to zero:
w + 7 = 0 or w - 4 = 0
Solving for "w":
w = -7 or w = 4
Since the width of the rectangle cannot be negative, we discard w = -7. Therefore, the width of the rectangle is 4 feet.
Using this value, we can find the length by adding 3 feet:
Length = Width + 3 = 4 + 3 = 7 feet
So, the dimensions of the rectangle are 4 feet by 7 feet.