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Parallelogram PQRS has a perimeter of 24. PQ = 2QR and Angle

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Parallelogram PQRS has a perimeter of 24. PQ = 2QR and Angle [#permalink]  29 Oct 2017, 04:02
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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?

[Reveal] Spoiler: OA
17.0

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Re: Parallelogram PQRS has a perimeter of 24. PQ = 2QR and Angle [#permalink]  15 Apr 2019, 08:30
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Hi Bunnel
Could you please post the solution for this?

Thanks so much!
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Re: Parallelogram PQRS has a perimeter of 24. PQ = 2QR and Angle [#permalink]  16 Apr 2019, 02:11
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PQRS has perimeter of 24. Consider x to be a side, then other side can be 2x.
3x*2=24, x=4.
So sides are 4 and 8.
One angle is 120, means other angle is 60.
Take sine of 60, which gives perpendicular like this..
sine of 60=p/4
thus, p=root3*2
find area as, A=p*8, (base times perpendicular height)
Area of another parallelogram is half, so it should be p*4..................(1)
Similarly using trigonometry find perpendicular height of smaller parallelogram, it should be x root3/2. (Where x is a smaller side and 2x would be larger)
now this height should be multiplied with '2x' to get area of smaller parallelogram..
Compare this to (1)...
Get x, it should be equal to 2 root2, so the other side will be 4 root2..
Then perimeter would be 2 root2 + 4 root2 + 2 root2 + 4 root2..
that is, 12 root2.. = 12*1.41 = 16.9
For nearest 0,1...it should be 17.0

( I am new here, this is my first post, sorry if this is not the way the answered should be presented. )
Thank you.
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Re: Parallelogram PQRS has a perimeter of 24. PQ = 2QR and Angle [#permalink]  16 Apr 2019, 03:06
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Welcome on board.

It is fine the way you have formatted.

Could be a bit better but is not the problem. Here we are to learn.

As long as you follow the rules for posting we are always eager to help the students.

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Re: Parallelogram PQRS has a perimeter of 24. PQ = 2QR and Angle [#permalink]  16 Apr 2019, 08:41
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]  16 Apr 2019, 10:08
Expert's post
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]  16 Apr 2019, 10:10
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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] 16 Apr 2019, 10:10
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