Final answer:
The maximum area of a rectangle with a perimeter of 12 meters is attained when the rectangle is a square with sides 3 meters long, yielding 9 m². However, since 9 m² is not an option and we're looking for the approximate maximum, the closest correct option from those provided is d) 8 m².
Step-by-step explanation:
The student's question pertains to finding the approximate maximum area of a rectangle given the perimeter is 12 meters. To maximize the area of a rectangle with a fixed perimeter, the rectangle must be a square (where all sides are equal).
For a square, the perimeter P is given by P = 4s, where s is the length of one side. Since we have a perimeter of 12 meters, we can divide this by 4 to find the length of one side, yielding s = 12/4 = 3 meters. The area of a square is then given by A = s². Therefore, the area of this square would be 3² = 9 square meters. However, as 9 square meters is not an option provided in the question, we must conclude that the 'rectangle' implied is not a square but likely has sides of different lengths.
To explore other possibilities for maximizing the area, we can see what happens with different length-width combinations while keeping the perimeter constant. For instance, if we had a rectangle with lengths 5 meters and 1 meter, the perimeter would be 5 + 5 + 1 + 1 = 12 meters, and the area would be 5 × 1 = 5 square meters. Similarly, if we tried lengths 4 meters and 2 meters, we would get an area of 4 × 2 = 8 square meters.
Therefore, considering this logic and the options given, the answer with the maximum area is less than 9 m², and the closest option to that is 8 m². So, the correct option is d) 8 m².