∆ABC is an equilateral triangle.
∴ AB = BC = AC = 6cm [Sides of an equilateral triangle]
pc = 1/3 BC [Given]
= 1/3 (6)
∴ PC = 2cm
In ∆ADC,
∠D = 90° [Construction]
∠C = 60° [Angle of an equilateral triangle]
∠DAC = 30° [Remaining angle of a triangle]
∴ ∆ ADC is a 30° – 60° – 90° triangle.
∴ AD = √3/2 AC [Side opposite to 60°]
∴ AD = √3/2 (6)
∴ AD = 3 √3 cm
CD = 1/2 AC [Side opposite to 30°]
∴ CD = 1/2 (6)
∴ CD = 3 cm
Now DP + PC = CD [D – P – C]
∴ DP + 2 = 3
∴ DP = 1 cm
In ∆ADP,
∠ADP = 900
AP2 = AD2 + DP2 [Pythagoras theorem]
∴ AP2 = (3√3)2 + (1)
∴ AP2 = 9 × 3 + 1 = 27 + 1
∴ AP2 = 28
∴ AP = √28
∴ AP = \(\sqrt{4 \times 7}\)
∴ AP = 2√7 cm.