QP=24 cm
RS=11.25 cm
QS=18.75 cm
Explanation:
Given that TQ bisects <RTP
[tex]therefore <QTP=<QTR\\[/tex](1)
consider ΔRQS and ΔRPT
QS||PT,RP and RT are transversals
[tex]<SQT=<QTP[/tex](alternate angles)(2)
comparing (1) and (2)
[tex]<SQT=<STQ[/tex] and triangle SQT is isocelus
Therefore SQ=ST(sides opposite to equal angles in an isocelus triangle)
Therefore <RQS=<RPT(corresponding angles)
<RSQ=<RTP(corresponding angles)
therefore by AA criterion for similarity
ΔRQS~ΔRPT
According to the property of similar triangles
[tex]RQ/RP=RS/RT=QS/PT[/tex][tex]9/RP=X/30=QS/50\\9/RP=X/30=(30-X)/50\\X/30=(30-X)/50\\50X=30(30-X)\\50X=900-30X\\50X+30X=900\\80X=900\\X=900/80=11.25\\RS=11.25 cm\\QS=30-X=30-11.25\\QS=18.75 cm\\9/RP=X/30\\9/RP=11.25/30\\9*30/11.25=RP\\RP=24 cm[/tex]
