Since △ RCD and △ STD are similar, and vector RC = 2 vector ST,

Then length RC : length ST = 1 : 2

The ratio of their corresponding lengths is 1 : 2

Ratio of their areas = 1² : 2²
= 1 : 4


Now RD : AD = 1 : 3
So AD = 3 RD

Area of the△ RCD = ½ x RD x perpendicular height of △RCD

Area of parallelogram
= AD x perpendicular height of parallelogram

= 3 RD x perpendicular height of △RCD
(They have the same perpendicular height)

= 6 x ½ RD x perpendicular height of △RCD

= 6 x area of △RCD

= 6 x 4 x area of △STD

= 24 x area of △STD

Ratio is 1 : 24

Maybe unitary approaches are easier to the eye.

Let the area of triangle STD be 1 unit².

Since triangles RCD and STD are similar with a scaling factor of 2,

Area of triangle RCD
= Area of triangle STD x (scale factor)²
= 1 unit² x 2²
= 4 units²

Since AR : RD = 2 : 1, it follows that RD : AD = 1 : 3.

Triangles RCD and ACD have bases in the ratio 1 : 3, but they share the common perpendicular height. As such, their area ratio is 1 : 3.

Area of triangle ACD
= Area of triangle RCD x 3
= 4 units² x 3
= 12 units²

Triangle ACD is basically a parallelogram ABCD being cut into two equal halves.

Area of parallelogram ABCD
= Area of triangle ACD x 2
= 12 units² x 2
= 24 units²
= 24 x Area of triangle STD

Therefore, area of triangle STD : area of parallelogram ABCD is 1 : 24.