Question

1. in the diagram below, all points are collinear. ( pt = 20 ), ( qs = 6 ), and ( pq = qr = rs ). find the length of ( rp ).

(\bigcirc) a. 20
(\bigcirc) b. 3
(\bigcirc) c. 6
(\bigcirc) d. 17

1. in the diagram below, all points are collinear. ( pt = 20 ), ( qs = 6 ), and ( pq = qr = rs ). find the length of ( sp ).

(\bigcirc) a. 20
(\bigcirc) b. 3
(\bigcirc) c. 9
(\bigcirc) d. 17

Explanation:

Problem 1:

Step1: Analyze QS and equal segments

Given \( PQ = QR = RS \) and \( QS = 6 \). Since \( QS=QR + RS \) and \( QR = RS \), let \( PQ = QR = RS = x \). Then \( QS=x + x=2x = 6 \).

Step2: Solve for x

From \( 2x = 6 \), we get \( x=\frac{6}{2}=3 \).

Step3: Find RP

\( RP=PQ + QR \), and \( PQ = QR = 3 \), so \( RP = 3+3 = 6 \)? Wait, no, wait. Wait, \( PQ=QR=RS = 3 \). Wait, \( RP=PQ + QR \), so \( PQ = 3 \), \( QR = 3 \), so \( RP=3 + 3=6 \)? But wait, let's check again. Wait, \( QS=QR + RS=3 + 3 = 6 \), correct. Then \( PT = 20 \), but for \( RP \), \( RP=PQ + QR \), \( PQ = 3 \), \( QR = 3 \), so \( RP = 6 \)? But wait, the options have C as 6. Wait, maybe I made a mistake. Wait, no, let's re - express.

Wait, \( PQ = QR = RS \), let each be \( x \). \( QS=QR + RS=2x = 6\Rightarrow x = 3 \). Then \( RP=PQ + QR=3 + 3 = 6 \). So the length of \( RP \) is 6.

Step1: Find the length of each equal segment

As in problem 1, since \( PQ = QR = RS \) and \( QS = 6 \), let \( PQ = QR = RS=x \). Then \( QS=QR + RS = 2x=6\Rightarrow x = 3 \).

Step2: Calculate SP

\( SP=PQ + QR+RS \). Since \( PQ = QR = RS = 3 \), then \( SP=3 + 3+3=9 \).

Answer:

C. 6

Problem 2: