seg PA, seg QB, seg RC and seg SD are perpendicular to line AD. [Given]
∴ seg PA || seg QB || seg RC || seg SD (i) [Lines perpendicular to the same line are parallel to each other]
Let the value of PQ be x and that of QR be y.
PS = PQ + QS [P – Q – S]
∴ 280 – x + QS
∴ QS = 280 – x (ii)
Now, seg PA || seg QB || seg SD [From (i)]
∴ AB/BD = PQ/QS [Property of three parallel lines and their transversals]
∴ AB / (BC + CD) = [B – C – D]
∴ 60/(70 + 80) = x/(280 - x)
∴ 60/150 = x/(280 - x)
∴ 2/5 = x(280 - x)
∴ 5x = 2 (280 – x)
∴ 5x = 560 – 2x
∴ 7x = 560
∴ x = 560/7 = 80
∴ PQ = 80 units
QS = 280 – x [From (ii)]
= 280 – 80
= 200 units
But, QS = QR + RS [Q – R – S]
∴ 200 = y + RS
∴ RS = 200 – y (ii)
Now, seg QB || seg RC || seg SD [From (i)]
∴ BC/CD = QR/RS [Property of three parallel lines and their transversals]
∴ 70/80 = y/(200 - y)
∴ 7/8 = y/(200 - y)
∴ 8y = 7(200 – y)
∴ 8y = 1400 – 7y
∴ 15y = 1400
∴ y = 1400/15 = 280/3
∴ QR = 280/3 units
RS = 200 – 7 [From (iii)]
= 200 – 280/2
= (200 x 3 - 280)/3
= (600-280)/3
∴ RS = 320/3 units
