Elena89 wrote:In triangle ABC, point X is the midpoint of side AC and point Y is the midpoint of side BC. If point R is the midpoint of line segment XC and if point S is the midpoint of line segment YC, what is the area of triangular region RCS ?

(1) The area of triangular region ABX is 32.
(2) The length of one of the altitudes of triangle ABC is 8.

St(1)
(1) The area of triangular region ABX is 32.






Area of ABX = 1/2 * h/2 * a = 32
=> 1/2 * h * a = 64

RS is parellel to AB and therefore CRS and CAB are similar triangles
Area of RCS = 1/2 * (h/4) * (a/4) = 64/16 = 4

Sufficient

St(2)
(2) The length of one of the altitudes of triangle ABC is 8.

The altitude could be for any vertex A, B or C, we dont know

Not sufficient


Option A

riju, i have some doubts for the portion underlined

rijul007 wrote:St(1)
(1) The area of triangular region ABX is 32.




Area of ABX = 1/2 * h/2 * a = 32
=> 1/2 * h * a = 64

RS is parellel to AB and therefore CRS and CAB are similar triangles
Area of RCS = 1/2 * (h/4) * (a/4) = 64/16 = 4

how you know they are parallel, the question doesn't specify neither ABC as isosceles triangle nor explicitly states RS || AB

pemdas wrote:riju, i have some doubts for the portion underlined

rijul007 wrote:St(1)
(1) The area of triangular region ABX is 32.




Area of ABX = 1/2 * h/2 * a = 32
=> 1/2 * h * a = 64

RS is parellel to AB and therefore CRS and CAB are similar triangles
Area of RCS = 1/2 * (h/4) * (a/4) = 64/16 = 4

how you know they are parallel, the question doesn't specify neither ABC as isosceles triangle nor explicitly states RS || AB

It is a general property in a triangle
A line parellel to one of the sides in a triangle divide the other non parellel sides proportionately

In triangle ABC, point X is the midpoint of side AC and point Y is the midpoint of side BC.

Join XY
We know that XY divides the two sides proportionately.. hence XY is parellel to AB
hence CXY and CAB are similar

If point R is the midpoint of line segment XC and if point S is the midpoint of line segment YC

Similarly, CXY and CRS are similar

Hence, CRS and CAB are similar


CR/AC = RS/AB = (Altitude of CRS)/(Altitude of CAB) = 1/4


Is this clear enough?

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.

@rijul007

How you got that height of triangle XAB is h/2. Kindly explain the same.

kakz wrote:@rijul007

How you got that height of triangle XAB is h/2. Kindly explain the same.

Ca=h

X and Y are midpoints of CA and CB respectively. so they divide Ca in two. that is why it is h/2

no need to calculate height.
the median of a triangle always divides the triangle into two parts with same area.
so 1) is sufficient. using above concept we can calculate the area of that small piece.
2) not enough

hence A

user123321