The given figure shows ΔABC and ΔPQR with AB = QR.

What is the value of (AB + 5BC + 3AC − 2RP − 3PQ − QR)?

In ΔABC,

∠ABC + ∠DBC = 180^o  [Linear pair]

⇒ ∠ABC + 130^o = 180^o

⇒ ∠ABC = 50^o

In ΔPQR,

∠PQS + ∠PQR = 180^o   [Linear pair]

⇒ 110^o + ∠PQR = 180^o

⇒ ∠PQR = 70^o

Also in ΔPQR,

∠PQS = ∠QPR + ∠PRS   [Exterior angle property]

⇒ 110^o = 60^o + ∠PRS

⇒ ∠PRS = 110^o − 60^o = 50^o

Now in ΔABC and ΔQRP,

AB = QR   [Given]

∠ABC = ∠QRP = 50^o

∠CAB = ∠PQR = 70^o

So, ΔABC ≅ ΔQRP   [By ASA congruence rule]

∴ BC = RP and AC = PQ   [CPCT]

BC = RP = 10 cm

Now,

AB + 5BC + 3AC − 2RP − 3PQ − QR

= AB + (5×10) + 3AC − (2×10) − 3AC − AB

= 50 − 20

= 30 cm

Thus, the value of the given expression is 30 cm.