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Please help me!!! (Please add an explanation)

The length of a rectangle is 21yd^2, and the length of the rectangle is 1yd less than twice the width. Find the dimensions of the rectangle

Find the length and width.

Please help me!!! (Please add an explanation) The length of a rectangle is 21yd^2, and-example-1
User Myermian
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Let's start by using algebra to solve for the width and length of the rectangle.

Let x be the width of the rectangle. Then, we know that the length is 1 yard less than twice the width. We can write this as:

length = 2x - 1

We also know that the length of the rectangle is 21 square yards. We can write this as:

length x width = 21

Substituting the expression for length from the first equation into the second equation, we get:

(2x - 1) x x = 21

Simplifying the equation:

2x^2 - x - 21 = 0

Now we can use the quadratic formula to solve for x:

x = (-b ± sqrt(b^2 - 4ac)) / 2a

In this case, a = 2, b = -1, and c = -21. Substituting these values into the quadratic formula, we get:

x = (-(-1) ± sqrt((-1)^2 - 4(2)(-21))) / 2(2)

Simplifying:

x = (1 ± sqrt(169)) / 4

x = (1 ± 13) / 4

x = 3 or x = -7/2

Since the width of a rectangle cannot be negative, we can ignore the negative solution. Therefore, the width of the rectangle is 3 yards.

Using the expression for length from the first equation, we can find the length of the rectangle:

length = 2x - 1

length = 2(3) - 1

length = 5

Therefore, the dimensions of the rectangle are 3 yards by 5 yards.
User Rodney Hickman
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