Elena, there was a very useful discussion on this thread
https://www.beatthegmat.com/triangle-2-d ... tml#439088 which actually enforced posters to prove many concepts and demonstrate their application ways. I will be resuming some concepts touched on in the copied thread:
a) median of a triangle is a line segment joining a vertex to the midpoint of the opposing side
b) median divides one triangle into two equal parts (two equal triangles)
c) similar triangles have to be similar: IF their corresponding angles are equal (congruent) OR IF their corresponding sides are correspondingly proportional
d) with similar triangles, the Ratio of Squares of two triangles is directly proportional to the the squared ratio of any two corresponding sides of the similar triangles (also found in MGMAT Geometry Strategy Guide 4th edition, page 33) s/S=(side/Side)^2
So cutting this short, I will be applying here par.b) from the above listed and decide that triangle(ABX)=triangle(BCX), hence their squares are 32 and the square of triangle(ABC) is 64 (32*2). Also applying par.a) from the above, triangle(CRS) is similar to triangle(ABC) and their sides are in ratio 1/4:1. Applying par.d) we have s/S=(1/4:1)^2 or s/64=1/16 and s=4
Sufficient with st(1), as square of triangle(CRS) is 4.
