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# halp

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In the figure below, PQRS is a parallelogram of perimeter  24 and area 28. What is the perimeter of rectangle QOST? Include an explanation of how you solved the problem.

Apr 18, 2018

#1
+101228
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Sincs RS  =  5.....then PQ  also equals 5 since RS  is parallel to PQ

So using the perimeter.... PS  + QR  =  24 - 5 -  5  =   14

And since PS = QR  then

2OR  = 14      divide both sides by 2

QR  = 7

And the area of  PQRS  = ( Base * Height)   = (QR * Height)

So

28  =  7 * Height       divide both sides by 7

4   =  Height    =  ST

And by the Pythagorean Theorem

RT  = sqrt  [ RS^2  - ST^2]  =   sqrt [ 5^2  - 4^2]  = sqrt [ 25 - 16 ] = sqrt (9)  =   3

So....the base  of QOST  = QT =  QR + RT   =  7 + 3   =  10

So  the area of QOST  =

QT  * ST  =

10 * 4   =

40 units^2

Apr 18, 2018
#2
+1

Since the perimeter of PQRS = 24 and QP=RS = 5 + 5 = 10. Therefore PS=QR =[24 - 10] / 2 = 7

And since the area of PQRS = 28, therefore the Height =28/7 =4=ST, since area of PQRS = base x height.Therefore the right triangle RST is a 3, 4, 5 triangle with RT = 3.

RT + QR =3 + 7 =10 =QT. Therefore the perimeter of the rectangle QOST =2[4 +10] = 28.

CPhill: I think the young person wants the PERIMETER, not the Area.

Apr 18, 2018
edited by Guest  Apr 18, 2018
#3
+101228
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Sorry, guest....my bad.....thanx for the better eyesight....LOL!!!!

CPhill  Apr 19, 2018