The length of a rectangle is
3yd less than twice the width, and the area of the rectangle is 65ydsquare. Find the dimensions of the rectangle.
L = 2 w - 3
and
L w = 65 so L = 65/w
65/w = 2 w - 3
65 = 2 w^2 - 3 w
2 w^2 - 3w - 65 = 0
(w+5))(2w-13)
positive w = 13/2 = 6.5
etc
To find the dimensions of the rectangle, we can use the given information about the length, width, and area.
Let's represent the width of the rectangle as "w" yds.
According to the problem, the length of the rectangle is 3 yds less than twice the width. So, the length can be represented as "2w - 3" yds.
The formula for the area of a rectangle is given by length multiplied by width. We are given that the area of the rectangle is 65 sq. yds. So, we can set up the following equation:
(2w - 3) * w = 65
Now, we can solve this equation to find the dimensions of the rectangle.
Expanding the equation, we get:
2w^2 - 3w = 65
Rearranging the equation:
2w^2 - 3w - 65 = 0
To solve this quadratic equation, we can either factorize it or use the quadratic formula. Let's use the quadratic formula:
w = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, a = 2, b = -3, and c = -65.
Substituting these values into the quadratic formula:
w = (-(-3) ± √((-3)^2 - 4 * 2 * -65)) / (2 * 2)
Simplifying the equation:
w = (3 ± √(9 + 520)) / 4
w = (3 ± √(529)) / 4
w = (3 ± 23) / 4
Thus, there are two possible solutions for the width:
Solution 1:
w = (3 + 23) / 4 = 26 / 4 = 6.5 yds
Solution 2:
w = (3 - 23) / 4 = -20 / 4 = -5
Since the width cannot be negative, we can disregard Solution 2.
Therefore, the width of the rectangle is 6.5 yards.
To find the length, we can substitute the value of the width back into the expression we found earlier:
Length = 2w - 3 = 2 * 6.5 - 3 = 13 - 3 = 10 yards
So, the dimensions of the rectangle are width = 6.5 yards and length = 10 yards.