∆ ABC is an equilateral triangle. Point P is on base BC such that if AB = 6 cm find AP.

ABC be an equilateral triangle,

Point P is on base BC, such that

Let us construct AM perpendicular on side BC from A.

As, ABC is an equilateral triangle we have

AB = BC = CA = 6 cm

Also, we know that Perpendicular from a vertex to corresponding side in an equilateral triangle bisects the side

Now, In ΔACM, By Pythagoras theorem

(Hypotenuse)² = (base)² + (Perpendicular)²

⇒ CA² = CM² + AM²

⇒ (6)² = (3)² + AM²

⇒ 36 = 9 + AM²

⇒ AM² = 27 [1]

As,

We have,

CM - PC = PM

⇒ PM = 1 cm

Now, In right angled triangle AMP, By Pythagoras theorem

(AP)² = (AM)² + (PM)²

⇒ (AP)² = 27 + 1²

⇒ AP² = 28

⇒ AP = 2√7 cm