Geometry of Touching Circles

Let the three circles have radius $r = 5$ cm. Let their centers be $O_1$, $O_2$, and $O_3$. Since the circles touch each other externally and have equal radii, the distance between the centers of any two touching circles is $r + r = 2r$.

The centers $O_1$, $O_2$, and $O_3$ form the vertices of an equilateral triangle ($\triangle O_1O_2O_3$). The side length of this triangle is $s_{centers} = 2r = 2 \times 5 = 10$ cm.

Triangle PQR from Contact Points

P, Q, and R are given as the points of contact between the circles. These points lie on the line segments connecting the centers ($O_1O_2$, $O_2O_3$, $O_3O_1$). Because the circles have equal radii, the points of contact P, Q, and R are precisely the midpoints of the sides of the triangle $\triangle O_1O_2O_3$.

Calculating Area of Triangle PQR

The triangle PQR is formed by connecting the midpoints of the sides of $\triangle O_1O_2O_3$. By the midpoint theorem in geometry, the triangle connecting the midpoints of a larger triangle is itself an equilateral triangle, and its side length is exactly half the side length of the larger triangle.

Therefore, the side length of $\triangle PQR$, denoted as $s_{PQR}$, is: $s_{PQR} = \frac{1}{2} \times s_{centers} = \frac{1}{2} \times (2r) = r$.

Given the radius $r = 5$ cm, the side length of $\triangle PQR$ is $5$ cm.

The formula for the area of an equilateral triangle with side length $s$ is: $ \text{Area} = \frac{\sqrt{3}}{4} s^2 $

Substituting the side length $s_{PQR} = 5$ cm into the area formula: $ \text{Area}(\triangle PQR) = \frac{\sqrt{3}}{4} (5 \text{ cm})^2 $ $ \text{Area}(\triangle PQR) = \frac{\sqrt{3}}{4} \times 25 \text{ cm}^2 $ $ \text{Area}(\triangle PQR) = \frac{25\sqrt{3}}{4} \text{ sq. cm} $

This calculated area corresponds to Option 2.