Step 1
Find the value of TS
we know that
if PQ is parallel to RS. then triangles TRS and TPQ are similar
so
[tex]\frac{TR}{TP} =\frac{TS}{QT}[/tex]
solve for TS
[tex]TS =\frac{TR*QT}{TP}[/tex]
we have
[tex]RP=2\ cm\\TP=18\ cm\\QT=27\ cm[/tex]
[tex]TR=TP+RP\\TR=18+2=20\ cm[/tex]
substitute
[tex]TS =\frac{20*27}{18}[/tex]
[tex]TS =30\ cm[/tex]
Step 2
Find the value of SQ
we know that
[tex]SQ=TS-QT[/tex]
we have
[tex]TS =30\ cm[/tex]
[tex]QT=27\ cm[/tex]
substitute
[tex]SQ=30\ cm-27\ cm=3\ cm[/tex]
therefore
the answer is
the value of SQ is [tex]3\ cm[/tex]
