What percentage of the square's area is shaded?

Dear Sasindran , would u please elaborate more why these triangles are proportionate?

Sure.

We all know that AB and CD are parallel lines. When two lines intersect between parallel lines the triangles so formed by them are always proportional to each other. Also the distance of intersection from either of the lines are also proportional to the length of the parallel lines.

For our case lets consider the intersecting point as O. Here we have AB=4 and DE=1. Hence by the above theorem we have perpendicular Distance between O and line AB is proportional to the perpendicular distance between O and the line DE.

i.e., AB/DE = per dist between O and AB/per dist between O and DE

4/1=height of triangle ABO/height of triagle DEO
also we know that height of triangle AEO+height of triagle DEO=4 (divided at O in the ratio 4:1)

Hence height of triangle AEO=4/5=3.2
height of triangle DEO=1/5=0.8

Hence area of triangle AEO=1/2*4*3.2=6.4

And you know the remaining septwibowo

Hope i made clear!

Thanks

septwibowo

Although may find 1000s of resources I am providing one:

https://www.varsitytutors.com/hotmath/h ... rtionality

Thank you very much Sasindran for your generosity! Kudo for you!

Thank you drexxie ! I forgot this theorem, my bad!

well, this question is hardly a gmat question. The source is also unreliable. However, this question is really a tough question.

This is real compact and accurate answer. Thanks mate. Kudo for you!

The question is based on similar triangles though the biggest challenge here is to identify that they are similar. Once you do, the question is straight forward.

Say the triangles are ABO and DEO (say O is the point of intersection of the lines)
Angle BOA = Angle DOA (vertically opposite angles)
Angle ABD = Angle EDB (alternate angles on traversal cutting two parallel sides of the square)

The triangles are similar by AA.
All corresponding sides of a similar triangle are in the same ratio.

Bases AB and DE are in the ratio 4 : 1
Imagine altitudes from O on bases AB and DE. They will be in the ratio 4 : 1 too. But they will also add up to 4, the side of the square.
So altitude of ABO will be (4/5)*4

Area of triangle ABO = (1/2) * 4 * (4/5)*4 = 32/5

Area of square = 4 * 4 = 16

Area of ABO as % of area of square = (32/5)/16 * 100 = 40%

I know this is something based on similar triangles concept, but instead of figuring out the actual values I tried to approximate and logically guess the answer.
Let the intersecting point of lines BD and AE be O.
Now we know that,
Area of square = 4 * 4 = 16
Area of triangle ABD = 16/2 = 8
Now, the shaded region is a portion of the triangle ABD. So the ratio should be less than 1/2. Eliminate Option E
We are given that ED = 1. Considering Triangle ADO, the altitude should be a less than 1. So the are of triangle ADO = 1/2 * 4 * 1 = 2. In fact something less than 2.
So the are of shaded region is < 8 - 2 i.e., Area of shaded region < 6

So it should be between 8/16 and 6/16. i.e., between 50% and 37.5%. So A, B and E are out. Since the height is close to 1, we can safely choose Option C.

This is not a fool proof solution, but it will help to quickly work, take a logical guess and move on.

I didn't know that altitudes are also similar in two similar triangles. New learning. Thanks Karishma

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