The length of a rectangle is twice the width. The area is 242 yd^2. Find the length and the width of the rectangle.
The width would be? The length would be?
I am not sure if I am to take 242 and sqrt it to give me 58,564 and then divide by 4 or what. I am lost with this.
no, work out PsyDAG's equation.
2W*W = 242
W^2 = 121
W = √121 = 11
so W = 11
L = 2W = 22
Check:
Area = 11*22 = 242
L = 2W
L * W = 242
2W * W = 242
Solve for W, then L.
So the L=242 and the W=60.5
Or am I way off base with this?
Or would the L=484 and the W=242
6+6
The area of a rectangle is
99 ft2
, and the length of the rectangle is
7 ft
more than twice the width. Find the dimensions of the rectangle.
To find the length and width of the rectangle, you can set up an equation based on the given information.
Let's start by assigning variables to the width and length. Let's call the width "w" and the length "l".
From the problem statement, we know that the length is twice the width, which can be written as:
l = 2w
The area of a rectangle is given by the formula:
Area = length × width
In this case, the area is given as 242 yd^2. So we have:
242 = l × w
Now we can substitute the value of l from the first equation into the second equation:
242 = (2w) × w
Simplifying further, we have:
242 = 2w^2
To solve for w, divide both sides of the equation by 2:
121 = w^2
Now, take the square root of both sides to find the value of w:
w = √121
w = 11
So, the width of the rectangle is 11 yards.
To find the length, substitute the value of w into the first equation:
l = 2w
l = 2 × 11
l = 22
Therefore, the length of the rectangle is 22 yards.
To summarize:
The width of the rectangle is 11 yards, and the length is 22 yards.