Recall that the diagonals of a parallelogram bisect each other, therefore:
[tex]\begin{gathered} PT=TR, \\ QT=TS. \end{gathered}[/tex]
Substituting the given expressions for PT, TR, QT, and TS, we get:
[tex]\begin{gathered} a+4=2a, \\ b=2b-3. \end{gathered}[/tex]
Solving each equation for its corresponding variable, we get:
[tex]\begin{gathered} 4=2a-a=a, \\ 3+b=2b, \\ 3=2b-b=b. \end{gathered}[/tex]
With the above values, we get that:
[tex]\begin{gathered} PR=a+4+2a=4+4+8=16, \\ QS=b+2b-3=3+6-3=6. \end{gathered}[/tex]
Answer:
[tex]\begin{gathered} 16\text{ length of PR,} \\ 4\text{ value of a,} \\ 3\text{ value of b,} \\ 6\text{ length of QS.} \end{gathered}[/tex]
