SSC CGL Tier II 1 Dec 2016 Paper 1

Question : 79 of 100

Marks: +1, -0

Let G be the centroid of the equilateral triangle ABC of perimeter 24 cm. Then the length of AG is

Solution:

(2)

AB = BC = CA =

24/3

= 8 cm.
BD = DC = 4 cm.; AD ⊥ BC
In ΔABD,

{AD}=√{ {AB}^{2}- {BD}^{2}}=√{8^{2}-4^{2}}

=√{64-16}=√{48}=4 \√{3} {cm}

∴{AG}= {2}/{3} {AD}= {2}/{3} ×4√{3}

= {8}/{√{3}} \ {cm} .

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