Question

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

Answer 1

∆ABC is an equilateral triangle.

It is given that,

PC = ${\displaystyle \frac{1}{3}}$BC

PC = ${\displaystyle \frac{1}{3}\times 6}$

PC = 2 cm

∴ BP = BC − PC

= 6 − 2

= 4 cm

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

OC = ${\displaystyle \frac{\text{BC}}{2}}$

= ${\displaystyle \frac{6}{2}}$

= 3 cm

∴ OP = OC − PC

= 3 − 2

= 1 cm   ...(1)

According to Pythagoras theorem,

In ∆AOB,

AB² = AO² + OC²

∴ (6)² = AO² + (3)²

∴ 36 = AO2 + 9

∴ AO² = 36 − 9

∴ AO² = 27

∴ AO = ${\displaystyle \sqrt{27}}$

∴ AO = ${\displaystyle 3\sqrt{3}}$ cm   ...(2)

In ∆AOP,

AP² = AO² + OP²

∴ AP² = ${\displaystyle {\left(3\sqrt{3}\right)}^2+{\left(1\right)}^2}$

∴ AP² = 27 + 1

∴ AP² = 28

∴ AP = ${\displaystyle \sqrt{28}}$

∴ AP = ${\displaystyle 2\sqrt{7}}$ cm