∆ PQR is right angled triangle, …

It is given that, ∠Q = 2∠PRQ. In ΔPQR, ∠PQR + ∠PRQ + ∠RPQ = 180° ⇒ 2∠PRQ + ∠PRQ + 90° = 180° ⇒ 3∠PRQ = 180° − 90° = 90° ⇒ ∠PRQ = 90°/30=30° … (1) ∴ ∠PQR = 2 × 30° = 60° … (2) Now, extending QP to S such that QP = PS, and joining RS. P /|\ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / | 90° \ /______ |__________\ S Q R In ΔRPQ and ΔRPS: RP = RP (Common) ∠RPQ = ∠RPS = 90° PQ = PS (by construction) ∴ ΔRPQ ≅ ΔRPS (by SAS congruence rule) ⇒ RQ = RS and ∠PRQ = ∠PRS (CPCT) Now, ∠SRQ = ∠PRQ + ∠PRS = 2∠PRQ = 2 × 30° = 60° [from (1)] By CPCT, we obtain ∠PQR = ∠PSR ∴ ∠PSR = 60° [from (2)] Therefore, ΔQRS is an equilateral triangle. ⇒ QR = RS = SP = 10 cm ∴ SP = PQ = 10cm/2=5cm On applying Pythagoras theorem to ΔPQR, we obtain RP2 + PQ2 = QR2 ⇒ RP2 + (5 cm)2 = (10 cm)2 ⇒ RP2 + 25 cm2 = 100 cm2 ⇒ RP2 = 75 cm2 ⇒ RP =5√3cm ∴ Perimeter of ΔPQR = RP + PQ + QR =(5√3+5+10)cm = (5 × 1.73 + 15) cm = 23.65 cm Thus, the perimeter of ΔPQR is 23.65 cm. The correct answer is C.

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