Question

In an equilateral triangle ABC, whose length of each side is 3 cm, D is the point on BC such that BD = (1/2) CD. What is the length of AD?

• A. $\sqrt{5}$ cm
• B. $\sqrt{6}$ cm
• C. $\sqrt{7}$ cm
• D. $\sqrt{8}$ cm
• E. None of the above

Hint:

In equilateral triangles, always use median properties and Pythagoras theorem for distance problems.

Answer 1

The Correct Option is C

Step 1: Recall equilateral triangle properties.
In $\triangle ABC$, each side = 3 cm.
Median from A to BC (say AE) is also altitude.
So, AE = $\tfrac{\sqrt{3}}{2} \times 3 = \tfrac{3\sqrt{3}}{2}$.
Step 2: Position of point D.
Given BD = $\tfrac{1}{2}$ CD.
So, BD : CD = 1 : 2.
Since BC = 3, BD = 1 cm, CD = 2 cm.
Step 3: Locate DE.
BE = CE = 1.5 cm.
So, DE = BE – BD = 1.5 – 1 = 0.5 cm.
Step 4: Use right triangle ADE.
In $\triangle ADE$:
$AD^2 = AE^2 + DE^2$.
$= \left(\tfrac{3\sqrt{3}}{2}\right)^2 + (0.5)^2$.
$= \tfrac{27}{4} + \tfrac{1}{4} = \tfrac{28}{4} = 7$.
Thus, $AD = \sqrt{7}$. \[ \boxed{\sqrt{7} \text{cm}} \]