ΔABC is an isosceles triangle and AB = AC = 2a unit, BC = a unit. Draw AD ⊥ BC , and find the length of AD

Answer: Option B

According to question,
Given :

AB = AC = 2a
BC = a
AD ⊥ BC
In isosceles triangle perpendicular sides bisects the opposite side of the length
∴ BD = $$\frac{BC}{2}$$
   BD = $$\frac{a}{2}$$
In ΔADB using Pythagoras theorem
AB² = BD² + AD²
(2a)² = $${\left(\frac{a}{2}\right)}^2$$ + AD²
4a² = $$\frac{a^2}{4}$$ + AD²
AD² = 4a² - $$\frac{a^2}{4}$$
AD = $$\frac{\sqrt{15}}{2}$$ a units