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GRE Parallelogram Challenge Question

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?



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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