In an equilateral triangle ABC, whose length
of each side is 3cm, D is a point on BC such thatBD = ½ CD. What is the length
of AD?

Answer follows here:

A. √5cm

B. √6cm

C. √7cm

D. √8cm

E. None of the aboveAnswer follows here:

**Solution:**
Let E be the mid point of BC. Hence,
BE = CE = 3/2

Given that BD = ½ CD; BD+CD = BC =
3;

Solving these two, we get BD = 1; CD = 2;

DE = BE-BD = 3/2 – 1 = 1/2

“For equilateral triangle, median AE becomes altitude”.
Hence AE is perpendicular to BC.

Applying Pythagoras to the right triangle ACE,

AE

^{2}= AC^{2}- CE^{2}= 3^{2}- (3/2)^{2}=> AE^{2}= 9 - 9/4 = 27/4
Applying Pythagoras to the right triangle ADE,

AD

^{2}= DE^{2}+ AE^{2}=> AD^{2}= (1/2)^{2}+ 27/4 = 28/4 = 7
=> AD = √7

**Answer (C)**

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