For any perimeter, the greatest possible area will be yielded if the perimeter forms a CIRCLE.What is the approximate minimum length of a rope required to enclose an area of 154 square meters?
A. 154
B. 60
C. 57
D. 50
E. 4
We can PLUG IN THE ANSWERS, which represent the smallest approximate perimeter required to yield an area of 154.
Since the question stem asks for the smallest possible perimeter, start with the smallest answer choice.
When the correct answer choice is plugged in, the perimeter will form a circle with an approximate area of 154.
E: 44
Here, 2πr = 44, implying that r = 44/(2π) = 22/π ≈ 7.
If this perimeter form a circle, we get:
A = πr² = π(7²) ≈ (22/7)(7²) = 154.
Success!
The correct answer is E.
Variation of the problem above:
RULE:What is the approximate minimum length of a rope required to enclose a rectangular area of 154 square meters?
A. 154
B. 60
C. 57
D. 50
E. 44
If rectangle R has a perimeter of x units, then the greatest possible area will be yielded if R is a SQUARE with a side of length x/4.
Example: Let p = 40
If L=10 and W=10, then A = 10*10 = 100.
If L=11 and W=9, then A = 11*9 = 99.
If L=12 and W=8, then A = 12*8 = 96.
As the example above illustrates, the greatest possible area is yielded when L=W=10 and R is a SQUARE.
We can PLUG IN THE ANSWERS, which represent the smallest approximate perimeter that will yield an area of 154.
Since the question stem asks for the smallest possible perimeter, start with the smallest answer choice.
E: 44
Here, the greatest possible area will be yielded if this perimeter forms a square with a side of 11.
If s = 11, then A = s² = 11² = 121.
Since the greatest possible area is too small, eliminate E.
D: 50
Here, the greatest possible area will be yielded if this perimeter forms a square with a side of 50/4.
If s = 50/4 = 25/2, then A = s² = (25/2)² = 625/4 ≈ 156.
Success!
The correct answer is D.













