Question

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

Given :- 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 .

To Find :- find an expression, in terms of y, for the length of PA ?

Solution :-

given that, ∆TBP ~ ∆TAQ . so,

→ BP/AQ = TP/TQ

→ 2/6 = TP/TQ .

→ TP/TQ = (1/3) -------- Eqn.(1)

now, in ∆PMT and ∆PAQ , we have,

→ ∠PMT = ∠PAQ (90°)

→ ∠MPT = ∠APQ (common)

so,

→ ∆PMT ~ ∆PAQ { By AA similarity. }

then,

→ PM/PA = PT/PQ { By CPCT .}

→ PM/PA = PT/(PT + TQ)

using Eqn.(1) in RHS,

→ PM/PA = 1/(1 + 3)

→ PM/PA = (1/4)

since,

→ PM = y m .

therefore,

→ y/PA = 1/4

→ PA = 4y m (Ans.)

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