A wire in the form of a square encloses an area of 144 cm². How much area is enclosed if the same wire is bent in the form of a rectangle of length 16 cm ?

The correct answer is 128 cm².

Step 1: Find the side length of the square.
Let the side of the square be $s$.
The area of the square is given as 144 cm².

$s^2=144$

$s=\sqrt{144}=12\text{ cm}$

Step 2: Find the length of the wire (perimeter of the square).
The perimeter of the square is equal to the total length of the wire.

$\text{Perimeter of square}=4\times s$

$\text{Perimeter}=4\times 12=48\text{ cm}$

Step 3: Find the breadth of the rectangle.
When the same wire is bent into a rectangle, the perimeter of the rectangle will be equal to the total length of the wire (48 cm).
Let the length of the rectangle be $l=16\text{ cm}$ and the breadth be $b$.

$\text{Perimeter of rectangle}=2\times (l+b)$

$2\times (16+b)=48$

$16+b=24$

$b=24-16=8\text{ cm}$

Step 4: Calculate the area of the rectangle.
The area of the rectangle is given by:

$\text{Area}=l\times b$

$\text{Area}=16\times 8=128{\text{ cm}}^2$

Thus, the area enclosed by the rectangle is 128 cm².