The perimeter of a rectangle is equal to the circumference of a circle. If the breadth of the rectangle is 8 cm and its length is twice the radius of the circle, what is the area of the circle?

Let the radius of the circle be r cm.

Let the length of the rectangle be L cm.

Let the breadth of the rectangle be B cm.

Now,

Breadth of the rectangle: $B=8\text{cm}$

Length of the rectangle in terms of radius: $L=2r\text{cm}$

Perimeter of rectangle = Circumference of circle:

Perimeter of rectangle $=2(L+B)$

Circumference of circle $=2\pi r$

So, $2(L+B)=2\pi r$

Substitute $L=2r$ and $B=8$ into the perimeter equation:

$2(2r+8)=2\pi r$

$2r+8=\pi r$

$8=\pi r-2r$

$8=r(\pi -2)$

Using $\pi \approx 22/7$:

$8=r\left(22/7-2\right)$

$8=r\left(22/7-14\right)$

$8=r\left(8/7\right)$

$r=8\times 7\; /\; 8$

$r\; =7\text{cm}$

The area of a circle is given by the formula:

$A=\pi r{}^2$

$A=22/7\times (7){}^2$

$A=22/7\times 49$

$A=22\times 7$

$A=154\text{cm}{}^2$

The final answer is 154 cm^2​.