The length of a rectangle is twice the width. If the area is 64^2cm, find the dimensions of the rectangle. Use the table or a calculator if necessary. Round to the nearest hundredths.
I do not know how to set the problem up. Thank you for your help.
width ----- x
length ---- 2x
x(2x) = 64
2x^2 = 64
x^2 = 32
x = √32 = .....
To set up the problem, we can start by assigning variables to the dimensions of the rectangle. Let's say that the width of the rectangle is "w" cm. According to the problem, the length of the rectangle is twice the width, which means the length would be "2w" cm.
To find the dimensions of the rectangle, we need to use the area of the rectangle formula, which is:
Area = Length × Width
Plugging in the values we have:
64^2cm = (2w) × w
So, we now have the equation:
4096 = 2w^2
To solve for "w", we need to isolate the variable on one side of the equation. Divide both sides of the equation by 2:
4096/2 = w^2
2048 = w^2
Now we can take the square root of both sides to solve for "w":
√2048 = w
Using a calculator, we find that √2048 is approximately 45.25.
Therefore, the width of the rectangle is approximately 45.25 cm.
To find the length, we can multiply the width by 2:
Length = 2 × 45.25 cm = 90.5 cm
So, the dimensions of the rectangle are approximately 45.25 cm (width) by 90.5 cm (length).