Let's analyze the problem involving an equilateral triangle and a circle inscribed within it. We are given an equilateral triangle ABC with a side length of 12 cm. We need to find the length of the radius of the circle that is inscribed inside this triangle.

Understanding the Inscribed Circle in an Equilateral Triangle

An inscribed circle within a triangle is the largest possible circle that can be drawn inside the triangle. It touches all three sides of the triangle. The center of the inscribed circle is known as the incenter of the triangle. In an equilateral triangle, the incenter coincides with the centroid, circumcenter, and orthocenter.

The radius of the inscribed circle is often called the inradius. There is a direct relationship between the side length of an equilateral triangle and the radius of its inscribed circle.

Formula for Inradius of an Equilateral Triangle

For an equilateral triangle with side length 'a', the formula for the inradius 'r' is given by:

$$ r = \frac{a}{2\sqrt{3}} $$

Alternatively, the height (h) of an equilateral triangle with side 'a' is $h = \frac{a\sqrt{3}}{2}$. The inradius 'r' is one-third of the height:

$$ r = \frac{1}{3}h = \frac{1}{3} \times \frac{a\sqrt{3}}{2} = \frac{a\sqrt{3}}{6} $$

Let's check if these two formulas are equivalent:

$$ \frac{a}{2\sqrt{3}} = \frac{a}{2\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{a\sqrt{3}}{2 \times 3} = \frac{a\sqrt{3}}{6} $$

Yes, both formulas are equivalent. We can use either one for the calculation.

Calculating the Radius of the Inscribed Circle

Given the side length of the equilateral triangle ABC is $a = 12$ cm.

We use the formula for the inradius 'r':

$$ r = \frac{a}{2\sqrt{3}} $$

Substitute the value of 'a' = 12 cm:

$$ r = \frac{12}{2\sqrt{3}} $$

Simplify the expression:

$$ r = \frac{6}{\sqrt{3}} $$

To rationalize the denominator, multiply the numerator and the denominator by $\sqrt{3}$:

$$ r = \frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} $$

$$ r = \frac{6\sqrt{3}}{3} $$

Simplify the fraction:

$$ r = 2\sqrt{3} $$

So, the length of the radius of the circle inscribed in the equilateral triangle is $2\sqrt{3}$ cm.

Result and Conclusion

The calculated radius of the inscribed circle is $2\sqrt{3}$ cm. This corresponds to one of the given options.

Revision Table: Equilateral Triangle Properties

Additional Information on Triangle Centers

In any triangle, there are several important centers. Let's look at a few, especially in the context of an equilateral triangle:

• Incenter: The intersection point of the angle bisectors. It is the center of the inscribed circle.
• Circumcenter: The intersection point of the perpendicular bisectors of the sides. It is the center of the circumscribed circle (a circle passing through all three vertices).
• Centroid: The intersection point of the medians. It is the center of mass of the triangle. Medians are line segments from a vertex to the midpoint of the opposite side.
• Orthocenter: The intersection point of the altitudes. Altitudes are perpendicular lines from a vertex to the opposite side.

For a general triangle, these four centers are usually distinct. However, for a special triangle like an equilateral triangle, all four centers coincide at the same point. This point is located at a distance of $\frac{1}{3}$ of the height from the base (which is the inradius, r) and $\frac{2}{3}$ of the height from the vertex (which is the circumradius, R).

Since $h = r + R$, and $R = 2r$, we get $h = r + 2r = 3r$, so $r = h/3$ and $R = 2h/3$. Using $h = \frac{a\sqrt{3}}{2}$, we get $r = \frac{1}{3} \times \frac{a\sqrt{3}}{2} = \frac{a\sqrt{3}}{6}$ and $R = \frac{2}{3} \times \frac{a\sqrt{3}}{2} = \frac{a\sqrt{3}}{3}$. These match the formulas previously mentioned.