The length of a rectangle is 5 cm more than its width. What are the possible dimensions of the rectangle if the area must be at least 84 cm2? (2 marks)
L = W + 5
L * W ≥ 84
Substitute W+5 for L in the second equation and solve for W. Insert that value into the first equation to solve for L. Check by putting both values into the second equation.
To find the possible dimensions of the rectangle, we can use the given information about the relationship between the length and width, as well as the minimum required area of 84 cm².
Let's start by assigning variables to the width and length of the rectangle. We can let the width be represented by "w" and the length be represented by "l".
According to the problem, the length of the rectangle is 5 cm more than its width. This can be expressed as:
l = w + 5
The area of a rectangle is given by the formula:
Area = length × width
Since we want the area to be at least 84 cm², we can set up the following inequality:
Area ≥ 84
Substituting the expressions for length and width, we get:
(w + 5) × w ≥ 84
Expanding the equation, we obtain:
w² + 5w ≥ 84
Rearranging the equation and setting it equal to zero to solve for the possible values of "w", we get:
w² + 5w - 84 ≥ 0
Now, we solve this quadratic inequality by factoring or using the quadratic formula. The solution should provide the range of possible values for "w".