Question

In DPQR as shown, ∠PQS = ∠RQS and QS ⊥ PR. Find the value of x and y, if PQ = 3x + 1; QR = 5y - 2; PS = x + 1 and SR = y + 2.

Answer 1

In ΔPQS and ΔSQR,
QS = QS               ....[Common]
∠QSP = ∠QSR    ....[each = 90°]
∠PQS = ∠RQS    ....[given]
∴ ΔPQS ≅ ΔSQR  ....[By ASA criterion]
⇒ PS = RS
⇒ x + 1 = y + 2
⇒ x = y + 1        ....(i)
And PQ = SQ
⇒ 3x + 1 = 5y - 2
⇒ 3(y + 1) + 1 = 5y - 2  ....[From (i)]
⇒ 3y + 3 + 1 = 5y - 2
⇒ 3y + 4 = 5y - 2
⇒ 2y = 6
⇒ y = 3
Putting y = 3 in (i),
x = y + 1
= 3 + 1
= 4.