(a) \(\frac{1}{5}\)
Area (ΔPSX) = Area (PUS) – Area (SUX)
In ΔPXS, WY || SX
⇒ \(\frac{PY}{YX}\) = \(\frac{PW}{WS}\) = 1 (Given, PW = WS)
⇒ PY = YX
In ΔRUY, SX || RY = \(\frac{UX}{XY}\) = \(\frac{US}{SR}\) = \(\frac{1}{2}\)
(∵ QS = SR and QV = US)
= UX = \(\frac{1}{2}\) (XY)
∴ In ΔPUS, UX = \(\frac{1}{2}\)XY = \(\frac{1}{2}\) \(\big(\frac{1}{2}PX\big)\) = \(\frac{1}{4}\) PX
PU = UX + PX = \(\frac{1}{4}\) PX + PX = \(\frac{5}{4}\)PX
∴ \(\frac{\text{Area of ΔSUX}}{\text{Area of ΔPUS}}\) = \(\frac{1}{5}\)
Now Area of ΔPUS = \(\frac{1}{4}\)(Area of ΔPQR)
Area of ΔSUX = \(\frac{1}{4}\) x \(\frac{1}{5}\) x Area of ΔPQR
= \(\frac{1}{20}\)Area of ΔPQR
∴ \(\frac{\text{Area of ΔPSX}}{\text{Area of ΔPQR}}\) = \(\frac{\frac{1}{4}Area\,of\,ΔPQR-\frac{1}{20}Area\,of\,ΔPQR}{Area\,of\,ΔPQR}\)
= \(\frac{5-1}{20}\) = \(\frac{1}{5}\).