In right triangle triangle ABC, C is a right angle; DE is parallel to

Hi Bunuel
Do you have any plan to post the solution?

KarishmaB Could you please explain this one? Thank you in advance!!

When mid points of two sides are joined, the line is parallel to the third side and the smaller triangle obtained is similar to the large triangle.



So when D and E are joined, triangle CDE is similar to triangle CAB (sides and altitude ratios 1 : 2) so the ratio of their areas is 1 : 4.

Area of triangle CDE = (1/2) * CT * DE = x ....... (I)
So area of triangle ABC = 4x

Area of rectangle DEMP = DE * TN (DP and EM are perpendicular to the base because the shaded triangles are similar to triangle ABC)
Since TN = CT (because altitude ratios are 1 : 2 so CT : CN = 1:2)

Area of rectangle DEMP = DE * CT = 2x (from (I) above)

Since the shaded triangles are congruent, K is the mid point of PM and hence, area of shaded triangles is half the area of rectangle DEMP i.e. area of shaded triangles = x

Shaded area/ Area of ABC = x/4x = 1/4

Answer (B)

https://gmatclub.com/forum/download/fil ... 33743fdeed

Sol. Draw a line on AB from C. let’s assume it as h.
Area of the triangle ABC =1/2×AB×h
Since shaded triangles are similar to the triangle ABC, hence shaded triangles are also right angle triangle. Height of the shaded triangle is ½ of the height ‘h’.
Combined base of both shaded triangles = PQ = DE = ½ of AB (as per midpoint theorem of triangle. )
Base of one shaded triangle = ¼ of AB
Area of both of the Shaded triangles =2×1/2×1/4 AB×1/2h=1/8×AB×h
Fraction of area shaded =((1/8×AB×h ))/((1/2×AB×h) )=1/4
Hence correct answer is B.