Question

The area of the rectangle is 24cm^(2). The with is two less than the length. What is the length and width of the rectangle?

Answer 1

The length and width of a rectangle with an area of 24cm^2 and the width being two less than the length are found by solving the quadratic equation l(l - 2) = 24. The length is 6 cm and the width is 4 cm.

Step-by-step explanation:

The question asks to find the length and width of a rectangle given that its area is 24cm2 and the width is two less than the length. To solve for the dimensions of the rectangle, we can set up an equation using the formula for the area of a rectangle, A = l x w, where A is the area, l is the length, and w is the width.

Let's denote the length of the rectangle as 'l' and the width as 'l - 2'. The area is given as 24 cm2, so we can set up the following equation: l x (l - 2) = 24. To find the values of 'l', we need to solve this quadratic equation.

Steps to solve the equation:

1. Expand the equation: l2 - 2l = 24.
2. Move all terms to one side to set the equation to zero: l2 - 2l - 24 = 0.
3. Factor the quadratic equation: (l - 6)(l + 4) = 0.
4. Solve for 'l' by setting each factor equal to zero: l - 6 = 0 or l + 4 = 0.
5. Find the length 'l' by solving the above equations: l = 6 or l = -4.
6. Since a rectangle cannot have a negative length, we take the positive value 'l = 6'.
7. Substitute 'l = 6' back into the width expression 'l - 2' to find the width: w = 6 - 2 = 4 cm.

Therefore, the dimensions of the rectangle are length = 6 cm and width = 4 cm.