A circle is inscribed in a rhombus with diagonals $12 \mathrm{~cm}$ and $16 \mathrm{~cm}$. The ratio of the area of circle to the area of rhombus is

We notice there's a mismatch between the question and the reference solution provided. The question is about geometry (circle inscribed in rhombus), but the reference solution discusses functions. We'll solve the geometry problem correctly while following AfterBoards standards.

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For any rhombus, when we know both diagonals, the area formula is:

$\text{Area of rhombus} = \frac{d_1 \times d_2}{2}$

The diagonals of a rhombus bisect each other at right angles, creating four right triangles. Each triangle has legs of length $\frac{d_1}{2}$ and $\frac{d_2}{2}$.

Given: $d_1 = 12$ cm and $d_2 = 16$ cm

$\begin{aligned} \text{Area of rhombus} &= \frac{12 \times 16}{2} \\ &= \frac{192}{2} \\ &= 96 \;\text{cm}^2 \end{aligned}$

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To find the radius of the inscribed circle, we need the side length first.

Using the diagonal relationship in a rhombus:

$\text{Side}^2 = \left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2$

The diagonals divide the rhombus into four congruent right triangles, and the side is the hypotenuse of each triangle.

$\begin{aligned} \text{Side}^2 &= \left(\tfrac{12}{2}\right)^2 + \left(\tfrac{16}{2}\right)^2 \\ &= 6^2 + 8^2 \\ &= 36 + 64 \\ &= 100 \end{aligned}$

$\text{Side} = 10 \text{ cm}$

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For any quadrilateral with an inscribed circle:

$\text{Area} = \text{inradius} \times \text{semiperimeter}$

The inscribed circle touches all four sides, and when we connect the center to each vertex, we get four triangles, each with height equal to the inradius.

Semiperimeter = $\frac{4 \times 10}{2} = 20$ cm

$96 = r \times 20$

$r = \frac{96}{20} = 4.8 \text{ cm}$

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$\begin{aligned} \text{Area of circle} &= \pi r^2 \\ &= \pi \times (4.8)^2 \\ &= \pi \times 23.04 \\ &= 23.04\pi \;\text{cm}^2 \end{aligned}$

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$\text{Ratio} = \frac{\text{Area of circle}}{\text{Area of rhombus}} = \frac{23.04\pi}{96}$

$\begin{aligned} \dfrac{23.04\pi}{96} &= \dfrac{23.04}{96}\pi \\ &= 0.24\pi \end{aligned}$

Converting to exact form:

$\frac{23.04}{96} = \frac{2304}{9600} = \frac{6}{25}$

Therefore, the ratio is $\frac{6\pi}{25}$

This makes sense because the circle is inscribed (smaller than the rhombus), and the ratio should be less than 1, which $\frac{6\pi}{25} \approx 0.75$ satisfies.