In an equilateral triangle ABC, whose length of each side is 3 cm, D is the point on BC such that BD = ½ CD. What is the length of AD?
Given : AB = AC = BC = 3 cm and BD = $$\frac{1}{2}$$ CD
AE is median.
To find : $$AD = ?$$
Solution : BD + CD = 3
=> $$BD + 2BD = 3BD = 3$$
=> $$BD = \frac{3}{3} = 1$$ cm
Also, since AE is media => $$BE = CE = \frac{3}{2}$$ cm
=> $$DE = BE - DE = \frac{3}{2} - 1 = \frac{1}{2}$$ cm
Also, AE = $$\frac{\sqrt{3}}{2} a = \frac{3 \sqrt{3}}{2}$$ cm
In $$\triangle$$ ADE
=> $$(AD)^2 = (AE)^2 + (DE)^2$$
=> $$(AD)^2 = (\frac{3 \sqrt{3}}{2})^2 + (\frac{1}{2})^2$$
=> $$(AD)^2 = \frac{27}{4} + \frac{1}{4} = \frac{28}{4}$$
=> $$AD = \sqrt{7}$$ cm