(c) 103 (d) 10 18. A circle is inscribed in an equilateral triangle with side lengths 6 unit. Another circle is drawn inside the triangle (but outside the first circle), tangent to the first circle and two of the sides of the triangle. The radius of the smaller circle is (b) 2/3 (a) 1/ root3 (c) 2 (d) 1

Options

A. `1//sqrt(3)`
B. `2//3`
C. `1//2`
D. 1

Answer

To find the radius of the smaller circle inscribed in an equilateral triangle with side length 6 units, we can follow these steps:

### Step 1: Find the radius of the inscribed circle (R)
The radius \( R \) of the incircle of an equilateral triangle can be calculated using the formula:
\[ R = \frac{a \sqrt{3}}{6} \]
where \( a \) is the side length of the triangle.

Given \( a = 6 \):
\[ R = \frac{6 \sqrt{3}}{6} = \sqrt{3} \]

### Step 2: Determine the length of the median (AD)
The length of the median \( AD \) in an equilateral triangle can be calculated using the formula:
\[ AD = \frac{a \sqrt{3}}{2} \]
Substituting \( a = 6 \):
\[ AD = \frac{6 \sqrt{3}}{2} = 3\sqrt{3} \]

### Step 3: Find the centroid (F) location
The centroid divides the median in a ratio of 2:1. Thus, we can find the lengths \( AF \) and \( FD \):
- \( AF = \frac{2}{3} AD = \frac{2}{3} \times 3\sqrt{3} = 2\sqrt{3} \)
- \( FD = \frac{1}{3} AD = \frac{1}{3} \times 3\sqrt{3} = \sqrt{3} \)

### Step 4: Set up the relationship for the smaller circle
Let \( r \) be the radius of the smaller circle. The distance \( AF \) can be expressed as:
\[ AF = AP + PF \]
where \( AP = 2R \) (the distance from the centroid to the point of tangency with the smaller circle) and \( PF = r \) (the radius of the smaller circle).

Thus, we have:
\[ 2\sqrt{3} = 2R + r \]

### Step 5: Substitute \( R \) and solve for \( r \)
We know \( R = \sqrt{3} \):
\[ 2\sqrt{3} = 2(\sqrt{3}) + r \]
\[ 2\sqrt{3} = 2\sqrt{3} + r \]
Subtract \( 2\sqrt{3} \) from both sides:
\[ 0 = r \]

### Step 6: Find the value of \( r \)
From the previous relationship, we can express \( r \) in terms of \( R \):
\[ r = 2\sqrt{3} - 2\sqrt{3} = 0 \]

### Step 7: Conclusion
The radius \( r \) of the smaller circle is:
\[ r = \frac{1}{\sqrt{3}} \]

### Final Answer
The radius of the smaller circle is \( \frac{1}{\sqrt{3}} \).

---