Tuesday, 25 October 2011

Geometry-4 (XAT-2010)

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?
A. √5cm
B. √6cm
C. √7cm
D. √8cm
E. None of the above
Answer 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) 


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