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# If parallelogram PQRS is inscribed in rectangle ORQT, as shown above,

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Joined: 02 Sep 2009
Posts: 58464
If parallelogram PQRS is inscribed in rectangle ORQT, as shown above,  [#permalink]

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29 Aug 2018, 03:46
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Difficulty:

65% (hard)

Question Stats:

65% (02:25) correct 35% (02:52) wrong based on 52 sessions

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If parallelogram PQRS is inscribed in rectangle ORQT, as shown above, and the rectangle has a perimeter of 26, what is the perimeter of the parallelogram?

(1) Area of parallelogram PQRS is 24.
(2) Area of the triangle RST is 6.

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image012.jpg [ 5.2 KiB | Viewed 580 times ]

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If parallelogram PQRS is inscribed in rectangle ORQT, as shown above,  [#permalink]

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29 Aug 2018, 05:39
Bunuel wrote:

If parallelogram PQRS is inscribed in rectangle ORQT, as shown above, and the rectangle has a perimeter of 26, what is the perimeter of the parallelogram?

(1) Area of parallelogram PQRS is 24.
(2) Area of the triangle RST is 6.

Attachment:
image012.jpg

Given, Perimeter of rectangle=2(OQ+QT)=26 or. OQ+QT=13-------(a)

Question stem:- what is the perimeter of the parallelogram?
We need the measure of 2 adjacent sides of the parallelogram.

St1:- Area of parallelogram PQRS is 24.
Or, QS*height=24 or, QS*OQ=24---------------(b)
The measure of both QS and OQ are unknown.
Insufficient.

St2:- Area of the triangle RST is 6.
RST is a right angled triangle. So, area=1/2*ST*RT=1/2*(QT-QS)*OQ=6 Or, (QT-QS)*OQ=12-------------(c)

We don't know the measure of QT.
hence insufficient to determine QS and RS.

Combined, from (c), we have
QT*OQ-QS*OQ=12
Or,QT*OQ-24=12 (from (b), QS*OQ=24)
Or, QT*OQ=36-----------(d)
From (a) and (d), we have QT=9 and OQ=4=RT (QT=4 and OQ=9 are invalid)
Now, 1/2*RT*ST=6 Or, ST=3. hence SR=5 (using Pythagorean property)
QS=QT-ST=9- 3=6

Perimeter of parallelogram can be calculated as we know the measure of adjacent sides.

Sufficient.

Ans. (C)
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If parallelogram PQRS is inscribed in rectangle ORQT, as shown above,   [#permalink] 29 Aug 2018, 05:39
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