A circle is inscribed in a rhombus with diagonals 12 cm and 16 cm. The ratio of the area of circle to the area of rhombus is

Let the diagonals of the rhombus be $d_1=12\text{cm}$ and $d_2=16\text{cm}$.

The area of the rhombus is given by:
${\text{Area}}_{\text{rhombus}}=\frac{1}{2}\times d_1\times d_2=\frac{1}{2}\times 12\times 16=96{\text{cm}}^2$

The diagonals of a rhombus bisect each other at right angles. The half-lengths of the diagonals are 6 cm and 8 cm. Thus, the side length a of the rhombus is:
$a=\sqrt{6^2+8^2}=10\text{cm}$

Let r be the radius of the inscribed circle. The radius of the inscribed circle in a rhombus can be found using the area and side length:
${\text{Area}}_{\text{rhombus}}=\text{base}\times \text{height}=a\times \left(2r\right)$
$96=10\times 2r\Rightarrow r=4.8\text{cm}=\frac{24}{5}\text{cm}$

The area of the inscribed circle is:
${\text{Area}}_{\text{circle}}=\pi r^2=\pi {\left(\frac{24}{5}\right)}^2=\frac{576\pi }{25}{\text{cm}}^2$

The ratio of the area of the circle to the area of the rhombus is:
$\text{Ratio}=\frac{\frac{576\pi }{25}}{96}=\frac{576\pi }{25\times 96}=\frac{6\pi }{25}$