∆ABC is an equilateral triangle. Point P is on base BC such that PC

1.	∆ABC is an equilateral triangle. Point P is on base BC such that PC = 1/3 BC, if AB = 6 cm find AP.Given: ∆ABC is an equilateral triangle.PC = 1/3 BC, AB = 6 cm.To find: APConsttuction: Draw seg AD ⊥ seg BC, B – D – C.
Answer» ∆ABC is an equilateral triangle. ∴ AB = BC = AC = 6cm [Sides of an equilateral triangle] pc = 1/3 BC [Given] = 1/3 (6) ∴ PC = 2cm In ∆ADC, ∠D = 90° [Construction] ∠C = 60° [Angle of an equilateral triangle] ∠DAC = 30° [Remaining angle of a triangle] ∴ ∆ ADC is a 30° – 60° – 90° triangle. ∴ AD = √3/2 AC [Side opposite to 60°] ∴ AD = √3/2 (6) ∴ AD = 3 √3 cm CD = 1/2 AC [Side opposite to 30°] ∴ CD = 1/2 (6) ∴ CD = 3 cm Now DP + PC = CD [D – P – C] ∴ DP + 2 = 3 ∴ DP = 1 cm In ∆ADP, ∠ADP = 900 AP² = AD² + DP² [Pythagoras theorem] ∴ AP² = (3√3)² + (1) ∴ AP² = 9 × 3 + 1 = 27 + 1 ∴ AP² = 28 ∴ AP = √28 ∴ AP = \(\sqrt{4 \times 7}\) ∴ AP = 2√7 cm.

∆ABC is an equilateral triangle. Point P is on base BC such that PC = 1/3 BC, if AB = 6 cm find AP.Given: ∆ABC is an equilateral triangle.PC = 1/3 BC, AB = 6 cm.To find: APConsttuction: Draw seg AD ⊥ seg BC, B – D – C.

Answer»

∆ABC is an equilateral triangle.

∴ AB = BC = AC = 6cm [Sides of an equilateral triangle]

pc = 1/3 BC [Given]

= 1/3 (6)

∴ PC = 2cm

In ∆ADC,

∠D = 90° [Construction]

∠C = 60° [Angle of an equilateral triangle]

∠DAC = 30° [Remaining angle of a triangle]

∴ ∆ ADC is a 30° – 60° – 90° triangle.

∴ AD = √3/2 AC [Side opposite to 60°]

∴ AD = √3/2 (6)

∴ AD = 3 √3 cm

CD = 1/2 AC [Side opposite to 30°]

∴ CD = 1/2 (6)

∴ CD = 3 cm

Now DP + PC = CD [D – P – C]

∴ DP + 2 = 3

∴ DP = 1 cm

In ∆ADP,

∠ADP = 900

AP² = AD² + DP² [Pythagoras theorem]

∴ AP² = (3√3)² + (1)

∴ AP² = 9 × 3 + 1 = 27 + 1

∴ AP² = 28

∴ AP = √28

∴ AP = \(\sqrt{4 \times 7}\)

∴ AP = 2√7 cm.