In the picture, quadrilateral ABCD is a parallelogram and

Dear Mike.. What is the likelihood of such a question on the GMAT. The more I see Kaplan questions, the more I feel the questions can be extremely hard. Whereas the questions on GMATPREP seem to be much simpler than this, No?

I personal think it would be on GMAT, but will be a 700 or 800 question. Calculation is straight forward. The only thing you need to recognize is that they both share the same triangle and if a triangle has the same height and width as a parallelogram thats not a trapezoid; then the triangle will always be 1/2 the area of the parallelogram. This is due to the simple mathematical equation to calculate the both of them.

Just my Personal opinion.

Dear omerrauf

I would say a question like this ---- a question that hinges on a relatively obscure geometry theorem that one probably would have to prove from scratch to answer the question ---- is something far harder than what they would put on the GMAT. Any GMAT math question, no matter how challenging, is something that someone facile with math would be able to solve in under a minute. If you've never seen this theorem, there's virtually no way that you will derive the full geometry proof in under a minute, unless you operate at Isaac Newton level. The GMAT doesn't expect that, even on 800 level questions. You don't have to have be Isaac Newton to answer the hardest questions.

That's my take on it. I am not as familiar with Kaplan questions overall, I am not qualified to make a statement about them. I know that Magoosh has a few hundred math questions, all appropriate difficulty for the GMAT, and each followed but its own video solution. The link above will give you a sample.

Mike

Dear Catennacio
That was a brilliant approach. Thank you for sharing that.
Mike

Just providing my 2 cents on the problem...

Theorem: Triangles between two parallel lines with same base have equal areas.

Even if you're not familiar with the above theorem, it's pretty intuitive from the area formula of the triangle.

In the figure, join EC.

Then from the above theorem, it's clear that area (tri DEC) = area (tri DBC) = \(\frac{1}{2}\)* area (ABCD) ---- (*)

Similarly, area (tri DEC) = area (tri DEF) = \(\frac{1}{2}\)* area (DEFG) ---- (**)

So, from (*) and (**), area (ABCD) = area (DEFG)

Clearly, (1) is sufficient and (2) is not, so answer is A.

This approach takes less than 2 minutes to solve. I think it is quite possible that similar questions are likely to be seen in GMAT at 700 level.

Got it intuitevly + elimination in 2 mins.

Stat.2 is clearly insuff, so eliminate B and D

Stat.1. Rectangle is a part of parallelogram or may be even equal

A,
sorry for my absence of discipline)

hi guys

here is another solution to this hard problem based on a different approach.

i hope you find the solution interesting and easy.

the diagram in the pdf is self explanatory.

Once we understand the diagram, the solution looks so easy.




Took me 15 minutes to figure out the approach.

Press kudos if you like the solution.

Please find the attached file for explanation

Answer: option A

Awesome explanation Gmat Insight!!! Thank you so much..

Somehow I came to know that in any parallelogram, if a triangle is formed joining two vertexes of the parallelogram and a point of opposite side of the vertexes, then the area of resulting triangle is half of the area of the parallelogram. Need confirmation from expert.

Draw a line EC.

Now in DEFG rectangle, Area of DEC = Area of \(\frac{1}{2}\) DEFG
Again in ABCD parallelogram, Area of DEC = Area of \(\frac{1}{2}\) ABCD

∴ Area of \(\frac{1}{2}\) DEFG = Area of \(\frac{1}{2}\) ABCD => Area of DEFG = Area of ABCD.

So, A is sufficient.

But how are we assuming that EC is perpendicular to DC? Please help.

Hi VeritasKarishma and GMATinsight, can you provide a ressponse to this question

Solution:

To solve this problem, we need to know the following fact:

If a triangle is inscribed in a parallelogram (or a rectangle) such that a side of the triangle coincides with a side of the parallelogram (or the rectangle) and a vertex not on of this side of the triangle is on the opposite side of the parallelogram (or the rectangle), then the area of the triangle is exactly half of the area of the parallelogram (or the rectangle).

With this fact in mind, we can draw EC and create triangle ECD. We see that the area of triangle ECD is exactly half the area of parallelogram ABCD since it satisfies the fact above. Furthermore, the area of triangle ECD is also exactly half the area of rectangle DEFG since it also satisfies the fact above. Therefore, it must be true that parallelogram ABCD and rectangle DEFG have the same area (since they are both twice the area of triangle ECD). In other words, if we know the area of the rectangle, we know the area of the parallelogram (and vice versa).

Statement One Only:

The area of rectangle DEFG is 8√5.

Since the area of rectangle DEFG is 8√5, the area of parallelogram ABCD is also 8√5. Statement one alone is sufficient.

Statement Two Only:

Line AH, the altitude of parallelogram ABCD, is 5.

Since we don’t know the length of DC, the base of the parallelogram, we can’t determine its area. Statement two alone is sufficient.

Answer: A

I would like to share my thought process. I maybe doing something crazy, but please bare with me I would answer this question using pure geometry after 20 minutes think, but under exam conditions ...

I answered this question purely using logic (which might not be very convincing). First of all, B and D are clearly out. Because we can not find the area of the parallelogram using only the height.

But if we have the area of the rectangle, and if I was able to replace one of the touching points of the two figures, lets say point E slightly to the left, and keeping in mind that point C must remain the touching point, then for the figure to remain rectangle remaining 2 points should also be rearranged accordingly. So any movement of point E requires other points to relocate as well and indicates exact locations of those other points.

Next step is that when we slide point E all the way to the left until it forms 90 degrees on top of point D. In this case point G will match point C and point F will hang on 90 degree above it as well. Now we already have a rectangle for which the height is exactly same as for parallelogram, and the side DC is also equal, therefore their areas must also be equal.

Apart from this questions, I have come across several problems, in which position of one point obligates the other point(s) to be exactly somewhere, in order for the conditions to hold (being rectangle in this case). In most of the cases what this means is that the area of the figure remains the same wherever you relocate there points.

I know this is an old thread, but I would be glad to hear assessment of my thought process from the experts.