Let, S and T be two points on the side QR of ΔPQR such that PS = PT and ∠QPS = ∠TPR. If PQ = (3x + 1) cm, value of \(\frac{(x+y)}{2}\) cm is ?

9.5

If PS = PT

⇒ ∠PST = ∠PTS

Thin in ΔPST

⇒ ∠PST = ∠SPQ + ∠SQP

∠PTS = ∠TRP + ∠TPR

[∠SPQ = ∠TPR]  Given

⇒ ∠PQS = ∠PRT

Now, in ΔPQR

∠Q = ∠R  ∴ PQ = PR [opposite sides of equal angles]

⇒ PQ = PR

⇒ 3x + 1 = 34

x = 11

ΔPQS ≅ ΔPRT

∴ PQ = PR

PS = PT

∠QPS = ∠TPR

∴ So, QS is also equal to RT

QS = RT

a = y + 1

y = 8

x + y =19

\(\frac{x+y}{2}\) = \(\frac{19}{2}\) = 9.5 cm