Bunuel wrote:

E is the midpoint of AC in right triangle ABC shown above. If the area of ΔABC is 24, what is the area of ΔBED?

A. \(9\sqrt{3}\)

B. \(9\sqrt{2}\)

C. 12

D. 9

E. \(6\sqrt{\frac{2}{3}}\)

This is how I tried.

Since we are given that E is the midpoint of AC then AE and EC are same.

We know that when the two bases are same the areas of triangles are also same..i.e.e 24 = 12+12 for two triangles.

Now BEC triangle is 12..

In BEC = BED + EDC triangles = 12.. we can happily discard option C because BED has to be lesser than 12...

From options we can say that BED area given then we can apply our logic.

A) 9 *1.7 = 15.3 which is greater than 12...not correct

B) 9*1.4 = 12.6 > 12...not correct

D) can be

E) 6*1.4/1.7 ~ 4.6... we can say from the shaded region BED is greater than EDC and this can't be the option.

IMO option D.