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Parallelogram PQRS has a perimeter of 24. PQ = 2QR and Angle : Numeric Entry Question

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Re: Parallelogram PQRS has a perimeter of 24. PQ = 2QR and Angle [#permalink]
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Welcome on board. :)

It is fine the way you have formatted.

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Re: Parallelogram PQRS has a perimeter of 24. PQ = 2QR and Angle [#permalink]
Thank you for the response viktorIartav, but what if we do not want to use trigonometry? Is there another way?
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Re: Parallelogram PQRS has a perimeter of 24. PQ = 2QR and Angle [#permalink]
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Trigonometry is not contemplated any more for the GRE exam.
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Re: Parallelogram PQRS has a perimeter of 24. PQ = 2QR and Angle [#permalink]
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let us take the first parallelogram. PQRS
PQ = 2QR
Perimeter - 2(PQ+QR)=2(2QR+QR)=6QR=24......QR=4 and PQ=2*4=8
Now let us see the area of PQRS
drop a perpendicular from Q on PS -> QA
Now PQA is 30-60-90 triangle as angle PQT = 120-90=30..
so if PQ is 8, height or PA will be \(8\sqrt{3}/2=4\sqrt{3}\)
Thus area = PS*QA = \(4*4\sqrt{3}\)=\(16\sqrt{3}\)...

so area of parallelogram TQUV is \(16\sqrt{3}/2=8\sqrt{3}\)..
now let QU be a, so TQ=2a...
height of TQUV will be TQ*\(\sqrt{3}/2=a\sqrt{3}\), as TQA' is also 30-60-90.
thus its area = \(a*a\sqrt{3}=a^2\sqrt{3}=8\sqrt{3}\), so \(a=2\sqrt{2}\)..
thus perimeter is 2(a+2a)=6a=6*\(2\sqrt{2}=12*1.414=17\)
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Re: Parallelogram PQRS has a perimeter of 24. PQ = 2QR and Angle [#permalink]
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