To solve this problem, we need to relate the properties of the inscribed circle (incircle) to the equilateral triangle. Here's the step-by-step working:

Step 1: Understand the relationship between the incircle and the equilateral triangle

For an equilateral triangle, the radius $r$ of the incircle is given by the formula: $r = \frac{s \sqrt{3}}{6}$ where $s$ is the side length of the equilateral triangle.

Step 2: Relate the circumference of the circle to the radius

The circumference $C$ of the incircle is given by: $C = 2 \pi r$ We're told that the circumference $C = 3$ cm.

Step 3: Solve for the radius $r$

Using the formula for circumference: $2 \pi r = 3$ $r = \frac{3}{2\pi}$

Step 4: Substitute the radius into the equation for $r$ in terms of $s$

Substitute $r = \frac{3}{2\pi}$ into the equation $r = \frac{s \sqrt{3}}{6}$: $\frac{3}{2\pi} = \frac{s \sqrt{3}}{6}$

Step 5: Solve for the side length $s$

Multiply both sides by 6 to get rid of the fraction: $\frac{18}{2\pi} = s \sqrt{3}$ $\frac{9}{\pi} = s \sqrt{3}$ Divide both sides by $\sqrt{3}$ to solve for $s$: $s = \frac{9}{\pi \sqrt{3}}$

Step 6: Rationalize the denominator

Multiply the numerator and the denominator by $\sqrt{3}$ to rationalize the denominator: $s = \frac{9 \sqrt{3}}{3\pi} = \frac{3 \sqrt{3}}{\pi}$

Step 7: Calculate the perimeter of the triangle

The perimeter $P$ of the equilateral triangle is given by: $P = 3s = 3 \times \frac{3 \sqrt{3}}{\pi} = \frac{9 \sqrt{3}}{\pi} \text{ cm}$

So, the exact perimeter of the triangle is $\frac{9 \sqrt{3}}{\pi}$ cm.

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Would you like further clarification on any step or additional details?

Related Questions:

1. How would the solution change if the triangle was not equilateral?
2. What is the relationship between the area of the triangle and its perimeter?
3. How can we calculate the area of the triangle with the given side length?
4. If the circumference was doubled, what would be the new perimeter?
5. How does the radius change if the side length of the triangle is increased?

Tip:

For equilateral triangles, the relationship between the side length and the incircle radius is particularly straightforward, making it easier to solve problems involving these figures.