Explanation :
For the perimeters of PQRS and PTUS,
①PS is common to both rectangles (shared side)
②TU has the same length as QR

The net difference is: (PQ + SR) - (PT + SU)
Since 10% of the rectangle PQRS is shaded, rectangle TQRU has 10% or 1/10 of PQRS's area.
Since their sides QR are the same, then this means the breadths TQ and UR must be 1/10 of the lengths PQ and SR
We can let TQ and UR be 1u each and PQ and SR be 10u each
Then this means that PT and SU are 9u each.
The overall difference is 8cm.
(10u + 10u) - (9u + 9u) = 8cm
20u - 18u = 8cm
2u = 8cm
1u = 8cm ÷ 2 = 4cm
(Notice that the difference is equal to the lengths of TQ + UR)
Finally, PT = 9u = 9 × 4cm = 36cm