#### Question

∆ABC is an equilateral triangle. Point P is on base BC such that PC = \[\frac{1}{3}\] BC, if AB = 6 cm find AP.

#### Solution

∆ABC is an equilateral triangle.

It is given that,

\[PC = \frac{1}{3}BC\]

\[ \Rightarrow PC = \frac{1}{3} \times 6\]

\[ \Rightarrow PC = 2 cm\]

\[ \Rightarrow BP = 4 cm\]

Since, ABC is an equilateral triangle, OA is the perpendicular bisector of BC.

∴ OC = 3 cm

⇒ OP = OC − PC

= 3 − 2

= 1 ...(1)

Now, According to Pythagoras theorem,

In ∆AOB,

\[{AB}^2 = {AO}^2 + {OB}^2 \]

\[ \Rightarrow \left( 6 \right)^2 = {AO}^2 + \left( 3 \right)^2 \]

\[ \Rightarrow 36 - 9 = {AO}^2 \]

\[ \Rightarrow {AO}^2 = 27\]

\[ \Rightarrow AO = 3\sqrt{3} cm . . . \left( 2 \right)\]

In ∆AOP,

\[{AP}^2 = {AO}^2 + {OP}^2 \]

\[ \Rightarrow {AP}^2 = \left( 3\sqrt{3} \right)^2 + \left( 1 \right)^2 \left( \text{From} \left( 1 \right) \text{and} \left( 2 \right) \right)\]

\[ \Rightarrow {AP}^2 = 27 + 1\]

\[ \Rightarrow {AP}^2 = 28\]

\[ \Rightarrow AP = 2\sqrt{7} cm\]

Hence, AP = 2\[\sqrt{7}\] cm.