Question

the segments are tangents to the circle. find the perimeter of jlno.
the perimeter of the polygon is
(simplify your answer)

Explanation:

Step1: Use the property of tangent segments

If two tangent segments are drawn from the same external point to a circle, then the lengths of the two tangent segments are equal.
So, \(JR = JK = 3\), \(LR = LM = 4\), \(NM = NP = 2\), \(QR = QP\)

Step2: Calculate the perimeter of \(JLNOQ\)

The perimeter \(P\) of \(JLNOQ\) is \(P=JL + LN+NQ + QJ\)
\(JL=JK + KL=3 + 4=7\)
\(LN = LM+MN=4 + 2=6\)
\(QJ=QR + RJ\) and \(QN=QP + PN\)
Since \(QR = QP\) and \(RJ = JK = 3\), \(PN = MN = 2\)
The perimeter \(P=(JK + KL)+(LM + MN)+(QP + PN)+(QR + RJ)\)
Substituting the equal - tangent values: \(P=(3 + 4)+(4 + 2)+(QR + 2)+(QR + 3)\)
But using the property of the perimeter of a polygon circumscribed about a circle \(P = 2\times(JK + LM+MN+RJ)\)
\(P = 2\times(3 + 4+2 + 3)\)

Step3: Compute the value

\(P=2\times(12)=24\)

Answer:

\(24\)