a rectangle has the length twice the side of this square and width three units less than the side of this square. if the two areas are equal, what are the squares dimensions (w * h)?
(2s)(s-3) = s^2
2s^2 - 6s = s^2
s^2 - 6s = 0
s(s-6) = 0
s=6
so, the square is 6x6
check: 12*3 = 36 = 6^2
To find the dimensions of the square, let's begin by assigning a variable to represent the side length of the square.
Let's call the side length of the square "x".
According to the given information:
- The length of the rectangle is twice the side length of the square, so its length is 2x.
- The width of the rectangle is three units less than the side length of the square, so its width is x - 3.
To find the area of the square, we square the side length:
Square's area = x * x = x^2
To find the area of the rectangle, we multiply its length and width:
Rectangle's area = (2x) * (x - 3) = 2x * (x - 3) = 2x^2 - 6x
Since the problem states that the two areas are equal, we can set up the equation:
x^2 = 2x^2 - 6x
Rearranging the equation:
2x^2 - 6x - x^2 = 0
Combining like terms:
x^2 - 6x = 0
Factoring out an x:
x(x - 6) = 0
Setting each factor equal to zero:
x = 0 (discarded since side length cannot be zero)
x - 6 = 0
Solving for x:
x = 6
Therefore, the side length (w and h) of the square is 6 units. Thus, the dimensions of the square are 6 units by 6 units (6 * 6).