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:
Solution:
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,
AE2 = AC2 - CE2 = 32
- (3/2)2 => AE2 = 9 - 9/4 = 27/4
Applying Pythagoras to the right triangle ADE,
AD2 = DE2 + AE2 =>
AD2 = (1/2)2 + 27/4 = 28/4 = 7
=> AD = √7
Answer (C)
No comments:
Post a Comment