The length of a rectangle is twice its with. The are is 128yd^2. Find the length and the Width
L = Length
W = With
A = Area
L = 2 W
A = L * W
A = 2 W * W
A = 2 W ^ 2
128 = 2 W ^ 2 Divide both sides with 2
64 = W ^ 2
W = sqrt ( 64 )
W = 8 yd
L = 2 W = 2 * 8 = 16 yd
A = L * W = 8 * 16 = 128 yd ^ 2
To solve this problem, we can use algebraic expressions. Let's assume that the width of the rectangle is "w" yards.
Given that the length of the rectangle is twice its width, we can express the length as "2w" yards.
To find the area of the rectangle, we multiply the length by the width. In this case, the area is given as 128 square yards. So, we can set up the equation:
Area = Length × Width
128 = (2w) × w
Simplifying the equation, we have:
128 = 2w^2
To solve for "w," we can divide both sides of the equation by 2:
2w^2/2 = 128/2
w^2 = 64
Next, we can take the square root of both sides to solve for "w":
√(w^2) = √64
w = ±8
Since the width cannot be negative in this context, we discard the negative value.
Thus, the width of the rectangle is 8 yards.
To find the length, we can substitute the width value into the expression for the length:
Length = 2w = 2 × 8 = 16 yards
Therefore, the length of the rectangle is 16 yards, and the width is 8 yards.