We have ΔPAB and ΔPQR

∠P = ∠P (Common)

∠PAB = ∠PQR (Corresponding angles)

Then, ΔPAB ~ ΔPQR (BY AA similarity)

So, (Corresponding parts of similar triangle area proportion)

Or,

Or, PB = x 6

Or, PB= 2cm

OR

We have, <ABC = 90° and BD Perpendicular AC

In Δ ABY and ΔBDC

<C = <C (Common)

<ABC = <BDC (Each 90° angles)

Then, Δ ABC ~ΔBDC (By AA Similarity)

So, (Corresponding parts of similar triangle area proportion)

Or

Or BC = 5.7/3.8 x 8.1

Or BC = 12.15cm