Question

qr || ps. find rs.
rs =

Explanation:

Step1: Use the basic proportionality theorem (Thales' theorem)

Since \( \overline{QR}\parallel\overline{PS}\), by the basic proportionality theorem in \(\triangle RTQ\), we have \(\frac{RS}{ST}=\frac{QP}{PT}\).
Let \(RS = x\), then \(ST=38\) (given), \(QP = 23\) (given), and \(PT = 69\) (given).
So, \(\frac{x}{38}=\frac{23}{69}\).

Step2: Solve the proportion for \(x\)

Cross - multiply: \(69x=23\times38\).
First, calculate \(23\times38 = 874\). Then \(x=\frac{874}{69}\).
Simplify \(\frac{874}{69}=\frac{23\times38}{23\times3}= \frac{38}{3}\approx12.67\) (This is wrong. Wait, actually, \(\frac{23}{69}=\frac{1}{3}\), so \(x=\frac{38}{3}\) is wrong. Wait, no, the correct proportion is \(\frac{RS}{RS + ST}=\frac{QP}{QT}\). Wait, no, the correct formula is \(\frac{RS}{ST}=\frac{QP}{PT}\). Wait, no, another way: Since \( \overline{QR}\parallel\overline{PS}\), \(\triangle RQP\sim\triangle TPS\) (by AA similarity, as \(\angle RQP=\angle TPS\) and \(\angle QRP=\angle TSP\) because of parallel lines). Then \(\frac{RS}{RT}=\frac{QP}{QT}\). But \(RT=RS + 38\) and \(QT=23 + 69=92\). So \(\frac{RS}{RS + 38}=\frac{23}{92}\).
Let \(RS=x\), then \(\frac{x}{x + 38}=\frac{23}{92}\). Cross - multiply: \(92x=23(x + 38)\).
Expand: \(92x=23x+874\).
Subtract \(23x\) from both sides: \(92x-23x=874\), \(69x = 874\), \(x = 14\).

Answer:

\(14\)