+68 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.
https://ibb.co/cRERSS
+129852 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
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.
