Question

circle o is inscribed in the given triangle. what is the perimeter of the triangle? 22 units 30 units 44 units 60 units

Explanation:

Step1: Recall tangent - segment property

If a circle is inscribed in a triangle, then the lengths of the two - tangent segments drawn from an external point to the circle are equal. Let the lengths of the tangent segments from the three vertices of the triangle to the points of tangency be \(x\), \(y\), and \(z\).
Let the side with length \(12\) be composed of two tangent - segments. Let the side with length \(6\) be composed of two tangent - segments and the side with length \(4\) be composed of two tangent - segments.
If we assume the segments as follows: Let the two segments on the side of length \(12\) be \(a\) and \(b\) (\(a + b=12\)), on the side of length \(6\) be \(b\) and \(c\) (\(b + c = 6\)) and on the side of length \(4\) be \(a\) and \(c\) (\(a + c=4\)).
We know that if a circle is inscribed in a triangle, the perimeter \(P\) of the triangle can be found using the formula \(P=2(s_1 + s_2 + s_3)\) where \(s_1\), \(s_2\), \(s_3\) are the lengths of the non - overlapping tangent segments from the vertices of the triangle to the points of tangency of the inscribed circle.
Let the three non - overlapping tangent segments be \(x\), \(y\), \(z\).
We know that if we consider the given lengths:
Let the lengths of the tangent segments be such that the perimeter \(P=(12 + 6+4)\times2\).

Step2: Calculate the perimeter

\[P=(12 + 6 + 4)\times2=(22)\times2 = 44\]

Answer:

44 units