The correct option is C (√2−1)√2
Given : Area (ΔAPQ) = Area (Trap. PBCQ)
We can see that, area (Δ APQ) = [Area (ΔABC) - Area (Trap. PBCQ)]
⇒2 Area (ΔAPQ) = Area (ΔABC)
⇒Area(ΔAPQ)Area(ΔABC)=12 ....(1)
Now, in ΔAPQ and ΔABC, we have ∠PAQ=∠BAC [Common]
∠APQ=∠ABC [since PQ || BC, corresponding angles are equal]
∴ΔAPQ∼ΔABC [by AA similarity criterion]
We know that the areas of similar triangles are proportinal to the squares of their corresponding sides.
∴Ar(ΔAPQ)Ar(ΔABC)=AP2AB2
⇒AP2AB2=12 [Using (i)]
→APAB=1√2
⇒AB=√2 (AB - PB)
⇒√2PB=(√2−1) AB
PBAB=(√2−1)√2