Question

5 numeric 1 point
if pq = qr, jk = 3x + 23 and lm = 9x - 19, find pk.
answer
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Explanation:

Step1: Use the property of equal - distance chords

Since \(PQ = QR\), chords \(JK\) and \(LM\) are equidistant from the center \(Q\) of the circle. By the property of a circle (chords equidistant from the center of a circle are equal in length), we have \(JK=LM\).
So, \(3x + 23=9x-19\).

Step2: Solve the equation for \(x\)

Subtract \(3x\) from both sides: \(23=9x - 3x-19\), which simplifies to \(23 = 6x-19\).
Add \(19\) to both sides: \(23 + 19=6x\), so \(42 = 6x\).
Divide both sides by \(6\): \(x=\frac{42}{6}=7\).

Step3: Find the length of \(JK\)

Substitute \(x = 7\) into the expression for \(JK\): \(JK=3x + 23\).
\(JK=3\times7 + 23=21 + 23=44\).

Step4: Use the property of perpendicular from the center to a chord

Since \(QO\perp JK\), \(PK=\frac{1}{2}JK\) (a perpendicular from the center of a circle to a chord bisects the chord).
\(PK=\frac{44}{2}=22\).

Answer:

\(22\)