In the given triangle PQR, QA and RB are the altitudes of sides PR and PQ respectively such that QA = RB.

If BQ = 3 cm, AP = 3 cm, and AQ = 4 cm, then what is the perimeter of ΔPQR?

It is given that QA = RB.

∴ RB = 4 cm   [Given, QA = 4 cm]

In ΔQBR, by Pythagoras theorem, we obtain

QR² = RB² + BQ²

⇒ QR² = (4)² + (3)² = 16 + 9 = 25

⇒ QR = 5 cm

In ΔBQR and ΔARQ,

QR = RQ   [Common]

∠RBQ = ∠QAR = 90^o

RB = QA   [Given]

So, ΔBQR ≅ ΔARQ   [RHS congruence rule]

∴ BQ = AR = 3 cm [By CPCT]

∠PQR = ∠PRQ   [By CPCT]

∴ PR = PQ   [Sides opposite to equal angles of a triangle are equal]

Now, PR = PA + AR = 3 cm + 3 cm = 6 cm

∴ PR = PQ = 6 cm

Perimeter of ΔPQR = PQ + QR + RP = (6 + 5 + 6) cm = 17 cm

Thus, the perimeter of ΔPQR is 17 cm.