This is a very high-difficulty problem!

This is a Numeric Entry question, so no answer choices– you’ll have to just type in your answer.

Give yourself 2 minutes to do this question, and let me know how it goes! 🙂

*Parallelogram PQRS has a perimeter of 24. PQ = 2QR and Angle PQR = 120 degrees. Parallelogram TQUV is formed by cutting out part of parallelogram PQRS so that angle TQU and PQR share the same angle. If TQUV is half the area of PRQS and TQ = 2QU, what is the perimeter of TQUV to nearest 0.1?*

Explanation:

This question can be solved relatively quickly if you remember that since the parallelograms are similar and one is twice the area of the other that any side of the large parallelogram is x√2 the length of the corresponding side of the small parallelogram (√2 x √2 = 2; which results in the twice the area).

Since we know that the sides of the large parallelogram are 4 and 8, we can derive the sides of the small parallelogram by √2, giving us 4/√2 = 2√2 and 8/√2 = 4√2. Therefore the perimeter of the small parallelogram (TQUV) is (4√2 + 2√2)2 = 16.97. Since the answer has to be rounded to the nearest .1, the answer is 17.0.

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Hey Chris………………..Waaadup???

Hi 🙂

Hey Chris, can we get a solution to this problem?

Hi Dan,

I just posted a solution to the problem. Hope that helps!

Hi Chris,

Are you sure that we need to find the perimeter of TQUV? Because according to my calculations, and if my calculations are right, the perimeter of TQUV is uncertain and have a range (though I’m not quite sure about the actual range). Is it possible for you to give out an illustration of the question? I think that T and U can go freely between the midpoint of PQ and P and the midpoint of QR and R respectively as long as the area of the parallelogram is half that of PQRS, making the perimeter of TQUV uncertain.

Looking forward to your reply.

Hi William,

You are definitely right! Thanks for catching that :). I’ve inserted into the question TQ = 2QR. That way we constrain where the points U and T fall along their respective line segments. Again, good catch!

Could that be an actual question on the GRE?

Hi David,

Well…it’s kind of tough. It depends. If you get the tough math section, a question of this difficulty/complexity could show up. But I think there would be very few questions that are this involved.

The thing is the question can be solved relatively quickly if you remember that since the parallelograms are similar and one is twice the area of the other that any side of the large parallelogram is x√2 the length of the corresponding side of the small parallelogram (√2 x √2 = 2; which results in the twice the area). Even the most difficult GRE math questions should be solvable in 2 minutes. So this passes muster :).

Hope that helps!