In the adjoining figure, seg PA , seg QB, seg RC and seg SD are perpendicular to line AD. AB = 60, BC = 70, CD = 80, PS = 280, then find PQ, QR and RS.

Explanation

seg PA, seg QB, seg RC and seg SD are perpendiculars to line AD.
Perpendiculars to same line are parallel to each other.
`:.` set `PA||` seg QB` ||` seg RC `||` seg SD…..1
Now seg AP `||` seg BQ `||` seg DS ….[From 1]
`:.` by property of three parllel lines and theri transversals, we get
`(SQ)/(PQ)=(DB)/(AB)`
By componendo we get
`(SQ+PQ)/(PQ)=(DB+AB)/(AB)`
`:.(PS)/(PQ)=(AD)/(AB)`...........(`P-Q-S` and `A-B-D`)
`:.280/(PQ)=(AB+BC+CD)/(AB)...........(A-B-C` and `B-C-D)`
`:.280/(PQ)=(60+70+80)/60`
`:.280/(PQ)=210/60`
`:.PQ=(280xx60)/210`
`:.PQ=80`
seg `AP||`seg `BQ||` seg CR.........[From 1]
`:.` by property of three parallel lines and their transversals, we get
`(PQ)/(QR)=(AB)/(BC)`
`:.80/(QR)=60/70`
`:.QR=(80xx70)/60`
`:.QR=280/3`
`PQ+QR+RS=PS...............(P-Q-R "and" Q-R-S)`
`:.80+280/3+RS=280`
`:.RS=280-80-280/3`
`:.RS=200-280/3`
`:.RS=(600-280)/3`
`:.RS=320/3`
`PQ=80,QR=280/3` and `RS=320/3`