The area of a rectangle is 243cm^2. The length is three times greater than the width. What are the dimensions of the rectangle?
A = LW
243 = W * 3W
243 = 3W^2
243/3 = W^2
81 = W^2
9 = W
Let's assume the width of the rectangle is "x" cm.
According to the given information, the length of the rectangle is three times greater than the width, so the length would be "3x" cm.
The formula for the area of a rectangle is length multiplied by width.
So, the area of the rectangle is given as:
x * 3x = 243 cm^2
Simplifying the equation, we have:
3x^2 = 243
Dividing both sides of the equation by 3, we get:
x^2 = 81
Taking the square root on both sides, we have:
x = 9
Therefore, the width of the rectangle is 9 cm, and the length is three times greater, which gives us:
Length = 3 * 9 = 27 cm
So, the dimensions of the rectangle are:
Width = 9 cm
Length = 27 cm.
To solve this problem, we need to set up an equation based on the given information.
Let's assume that the width of the rectangle is "w" cm. Since the length is three times greater than the width, the length would be "3w" cm.
Now, we can use the formula for the area of a rectangle, which is length multiplied by width: Area = Length * Width.
Given that the area of the rectangle is 243 cm^2, we can write the equation as follows:
243 = (3w) * w
Now, we can solve this equation for "w".
First, distribute the 3w:
243 = 3w^2
Next, divide both sides of the equation by 3:
81 = w^2
Now, take the square root of both sides to find the value of "w":
w = √(81)
w = 9
So, the width of the rectangle is 9 cm.
Since the length is three times greater, the length would be 3 * 9 = 27 cm.
Therefore, the dimensions of the rectangle are 9 cm (width) and 27 cm (length).