In the figure 1.76, 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.
Figure: A, B, C, D lie in that order on line AD; P, Q, R, S lie in that order on another transversal line PS.
Answer: PQ = 80, QR = \(\frac{280}{3}\), RS = \(\frac{320}{3}\)
Lines perpendicular to the same line are parallel, so PA ∥ QB ∥ RC ∥ SD. Parallel lines cut two transversals in proportional intercepts:
\[\frac{PQ}{AB} = \frac{QR}{BC} = \frac{RS}{CD}\]So PQ : QR : RS = 60 : 70 : 80 = 6 : 7 : 8, and PQ + QR + RS = PS = 280. Using the sum of the parts (6 + 7 + 8 = 21):
\[PQ = \frac{6}{21} \times 280 = 80\] \[QR = \frac{7}{21} \times 280 = \frac{280}{3}\] \[RS = \frac{8}{21} \times 280 = \frac{320}{3}\]Log in or create a free account to answer this question.
