Step-by-step explanation:
In ∆PQR and ∆TSU
[tex]\because\begin{cases}PQ\approx TS \\ QR\approx TU \angle \\ Q\approx \angle T\end{cases}[/tex]
[tex]\therefore{∆PQR\approx ∆TSU(\because SAS)}[/tex]
And
- <Q is greater than <T
- So SU will be less than PR(22cm)
