Geometric Setup: Three Touching Circles
Imagine three identical circles, each with radius r, are placed so that each one touches the other two. This arrangement forms a compact cluster.
The centers of these three small circles form the vertices of an equilateral triangle. The distance between the centers of any two touching circles is equal to the sum of their radii, which is \(r + r = 2r\). Therefore, the side length of the equilateral triangle formed by the centers is \(2r\).
The Circumscribing Circle
A larger circle is drawn around these three smaller circles such that it touches the outer edge of each of them. This is called the circumscribing circle.
- Let the radius of this larger, circumscribing circle be R.
- The center of the circumscribing circle coincides with the centroid of the equilateral triangle formed by the centers of the smaller circles.
Calculating the Radius (R) of the Circumscribing Circle
The distance from the centroid of an equilateral triangle to any of its vertices is given by the formula \(\frac{a}{\sqrt{3}}\), where a is the side length of the triangle.
In our case, the side length \(a = 2r\). So, the distance from the center of the large circle (the centroid) to the center of any small circle (a vertex) is:
\( \text{Distance} = \frac{2r}{\sqrt{3}} \)The radius R of the circumscribing circle is the distance from its center to the center of a small circle PLUS the radius r of that small circle (because the large circle touches the *outer edge* of the small circle).
\( R = (\text{Distance from center to center}) + r \) \( R = \frac{2r}{\sqrt{3}} + r \)Factoring out r:
\( R = r \left( \frac{2}{\sqrt{3}} + 1 \right) \)Combining the terms inside the parenthesis:
\( R = r \left( \frac{2 + \sqrt{3}}{\sqrt{3}} \right) \)Using the Given Area to Find R
The problem states that the area of the circumscribing circle is \(\frac{\pi(2+\sqrt{3})^2}{3}\) square cm.
We know the formula for the area of a circle is \(\pi R^2\). So, we can set up an equation:
\( \pi R^2 = \frac{\pi(2+\sqrt{3})^2}{3} \)Divide both sides by \(\pi\):
\( R^2 = \frac{(2+\sqrt{3})^2}{3} \)Take the square root of both sides:
\( R = \sqrt{\frac{(2+\sqrt{3})^2}{3}} \) \( R = \frac{2+\sqrt{3}}{\sqrt{3}} \)Finding the Radius (r) of the Smaller Circles
Now we have two expressions for R:
- \( R = r \left( \frac{2 + \sqrt{3}}{\sqrt{3}} \right) \) (from our geometric derivation)
- \( R = \frac{2+\sqrt{3}}{\sqrt{3}} \) (from the given area)
Let's equate these two expressions:
\( r \left( \frac{2 + \sqrt{3}}{\sqrt{3}} \right) = \frac{2+\sqrt{3}}{\sqrt{3}} \)To find r, we can see that both sides have the factor \(\frac{2+\sqrt{3}}{\sqrt{3}}\). Dividing both sides by this factor leaves us with:
\( r = 1 \)So, the radius of one of the smaller circles is 1 cm.
Conclusion
By analyzing the geometric relationship between the three touching identical circles and the larger circumscribing circle, and using the given area, we determined the radius of the smaller circles.
The radius of one of the smaller circles is 1 cm.
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