1.

In an equilateral triangle ABC, the side BC is trisected at D, then 9AD2 is .......(1) 7AB2 (2) 8BC2 (3) 4AC2(4) 3/2 AB2

Answer»

Correct option is (1) 7AB2

Let A(0, 0), B(2a, 0), C(a, √3 a),

Then D \((\frac{4a}{3},\frac{2\sqrt{3}}{3})\)

⇒ 9 AD2 = 28a2 

= 7 AB2



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