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.

Answer 1

∆ABC is an equilateral triangle.

It is given that,
$$\begin{gathered} \mathrm{PC}=\frac{1}{3}\mathrm{BC} \\ \Rightarrow \mathrm{PC}=\frac{1}{3}\times 6 \\ \Rightarrow \mathrm{PC}=2\mathrm{cm} \\ \Rightarrow \mathrm{BP}=4\mathrm{cm} \end{gathered}$$

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,

$$\begin{gathered} {\mathrm{AB}}^2={\mathrm{AO}}^2+{\mathrm{OB}}^2 \\ \Rightarrow {\left(6\right)}^2={\mathrm{AO}}^2+{\left(3\right)}^2 \\ \Rightarrow 36-9={\mathrm{AO}}^2 \\ \Rightarrow {\mathrm{AO}}^2=27 \\ \Rightarrow \mathrm{AO}=3\sqrt{3}\mathrm{cm}...\left(2\right) \end{gathered}$$

In ∆AOP,

$$\begin{gathered} {\mathrm{AP}}^2={\mathrm{AO}}^2+{\mathrm{OP}}^2 \\ \Rightarrow {\mathrm{AP}}^2={\left(3\sqrt{3}\right)}^2+{\left(1\right)}^2\left(\mathrm{From}\left(1\right)\mathrm{and}\left(2\right)\right) \\ \Rightarrow {\mathrm{AP}}^2=27+1 \\ \Rightarrow {\mathrm{AP}}^2=28 \\ \Rightarrow \mathrm{AP}=2\sqrt{7}\mathrm{cm} \end{gathered}$$

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