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 ?
Answer & explanation
Correct answer: option 2
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