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 cm^2? (2 marks)
To find the possible dimensions of the rectangle, we need to define some variables and set up an equation based on the given information.
Let's assume the width of the rectangle is "w" cm.
According to the given information, we know that the length of the rectangle is 5 cm more than its width. So, the length can be represented as "w + 5" cm.
The area of a rectangle is given by the formula: Area = length × width.
In this case, the area must be at least 84 cm². So we can set up the following inequality:
(w + 5) × w ≥ 84
Now, let's solve this inequality to find the possible values of width.
w² + 5w ≥ 84
Rearranging the terms:
w² + 5w - 84 ≥ 0
Now we have a quadratic inequality. We can solve it by factoring or using the quadratic formula. In this case, let's factor it:
(w - 7)(w + 12) ≥ 0
Now we have two factors, (w - 7) and (w + 12). To satisfy the inequality, either both factors must be positive or both factors must be negative.
Case 1: Both factors are positive:
w - 7 ≥ 0 ⇒ w ≥ 7
w + 12 ≥ 0 ⇒ w ≥ -12 (This inequality is always true and does not impose any restrictions.)
Case 2: Both factors are negative:
w - 7 ≤ 0 ⇒ w ≤ 7
w + 12 ≤ 0 ⇒ w ≤ -12
Since the width cannot be negative, we discard the solution w ≤ -12.
Therefore, the possible value(s) for the width of the rectangle is w ≥ 7 cm.
To find the corresponding length, we can substitute this value back into the equation for the length:
length = w + 5
length = 7 + 5
length = 12 cm
So, the possible dimensions of the rectangle are width ≥ 7 cm and length = 12 cm.