Question

∆ ABC is an equilateral triangle. P is the point on side BC such that PC = 1 / 3 BC. If AB = 6 cm, then find AP.

Answer 1

PC = 1 / 3 BC, AB = 6

∆u2009ABC is an equilateral triangle.

∴ AB = BC = AC = 6

∴ PC = 1 / 3 BC = 1 / 3 × 6 = 2

Draw Seg AD ⊥ Seg BC.

In ∆ DAC, ∠u2009ADC = 90˚, ∠ACB = 60˚ ∴ ∠DAC = 30˚

∴ DC = 1/2 × AC = 1/2 × 6 = 3 .........(30o ,60o ,90o theorem)

∴ DP = DC - PC = 3 - 2 = 1.

Now , AD = √3 / 2 × AC = √3 / 2 × 6 = 3 √3

In ∆u2009ADP

AP2 = AB2 + DP2 .............(Pythagoras theorem)

= (3 √3 )2 + 12

= 9 × 3 + 1

= 28

∴ AP = √28 = 2 √7 cm