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06 Dec 2018, 04:55
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
50% (03:12) correct
50%
(02:50)
wrong
based on 34
sessions
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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?
A. 15
B. 17
C. 18
D. 19
E. 20
A. 15
B. 17
C. 18
D. 19
E. 20
Archit3110
Major Poster
06 Dec 2018, 06:08
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Bunuel
chetan2u : I am not able to visualize this question on paper.. could you please look into the question..
06 Dec 2018, 06:36
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Archit3110
See the figure attached...
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*\(2\sqrt{2}=12*1.414=17\)
B
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