ABCD is a rectangle. P, Q and R are the midpoint of BC, CD and DA. The point S lies on the line QR in such a way that SR: QS = 1:3. The ratio of the area of triangle APS to area of rectangle ABCD is

First, let us construct the rectangle using the given information.

Let the length of the rectangle be 'l' and the breadth of the rectangle be 'b'.
Area of rectangle ABCD = lb. 
We have to find the area of the triangle APS.

We consider length as 4cm and breadth as 2cm. ( Answer is irrespective of length and breadth) 

In triangle QRD

RS/RQ=XS/DQ

1/4=XS/2cm

XS= 0.5 cm

SY= 4cm-0.5cm= 3.5cm

We can now find the areas of all the sections.

area (RDQ)= 1/2 x 1cm x 2cm= $$1cm^2$$

area (ABP) =1/2 x 1cm x 4cm = $$2cm^2$$

area (ARS) =1/2 x 1cm x 0.5cm = $$0.25cm^2$$

area (PSQC) = Area (PSZ)-Area(QCZ)= 1/2 x 2cm x 3.5cm  - 1/2 x 1cm x 2cm =

                       =   $$2.5cm^2$$

Area (ABCD)= 4cm x 2cm = $$8cm^2$$

Area (APS) = 8 -1 -2 - 0.25 - 2.5 = $$2.25cm^2$$ 

Area (APS) / Area (ABCD) = 9/32  =  36/128

Hence answer (A).