$\left.\begin{array}{l}\text { Rhombi. Properties and Characteristics } \\ \text { [Area of a Circle, Sector, and Segment] }\end{array}\right]$

A circle with a radius of 3 cm is inscribed in a rhombus, one of whose diagonals is 10 cm. Calculate the area of the part of the rhombus that is outside the circle. Will this area be greater than 9 cm²? (Justify your answer.)
Diagonals divide the rhombus into four equal right triangles.

## Solution

Let the inscribed circle in rhombus $ABCD$ with center $O$ have a radius $r=3$ and touch side $AB$ at point $M$, and diagonal $AC=10$. Then $OM$ is the height of the right triangle $AOB$ drawn from the vertex of the right angle. Therefore,

$$
AM=\sqrt{AO^2-OM^2}=\sqrt{25-9}=4, \quad AB=\frac{AO^2}{AM}=\frac{25}{4}
$$

Then

$$
S_{\mathrm{ABCD}}=4 \cdot S_{\triangle \mathrm{AOB}}=4 \cdot \frac{1}{2} \cdot AB \cdot OM=2 \cdot \frac{25}{4} \cdot 3=\frac{75}{2}
$$

Let $S_{1}$ be the area of the circle, and $S$ be the desired part of the area of the rhombus. Then

$$
S=S_{\mathrm{ABCD}}-S_{1}=\frac{75}{2}-\pi r^{2}=\frac{75}{2}-9 \pi
$$

Since $\pi \approx 3.14$, we have

$$
\frac{75}{2}-9 \pi \approx \frac{75}{2}-28.26=37.5-28.26=9.24 > 9
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_1b17058527caf5b5d058g-31.jpg?height=626&width=828&top_left_y=2214&top_left_x=615)

![](https://cdn.mathpix.com/cropped/2024_05_06_1b17058527caf5b5d058g-32.jpg?height=343&width=683&top_left_y=0&top_left_x=693)

## Answer

$\frac{75}{2}-9 \pi > 9$.