Question

please help.
Two ladders AB and PQ are resting against opposite walls of an alley. The ladders AB and PQ we 2m and 6m above the ground respectively and T is the point where the two ladders meet. Given that ∆TBP is similar to ∆TAQ, find an expression, in terms of y, for the length of PA​

Answer 1

Answer:

Step-by-step explanation:

∠PBA=∠BAQ(alternate angle)

∠BTP=∠QTA

∠BPT=∠TQA

$\frac{PT}{QT}$=$\frac{BT}{AT}$=$\frac{BP}{AQ}$=1/3

ΔPMT≈ΔPAQ and ΔAMT≈ΔAPB

PM/PA=PT/(PT+QT)=TM/QA------eq1

1/4=TM/6

TM=1.5m

let line of length y m intersect at line PB at O

then PO=TM=1.5m

TP=$\sqrt{1.5^{2}+y^{2} }$

PM=$\sqrt{TP^{2}-TM^{2} }$

PM=y

PM/PA=1/4(eq1)

PA=4y