In ΔABC, AB = BC = K, AC = $$\sqrt 2 $$ k, then ΔABC is a : – McqOptions

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1.	In ΔABC, AB = BC = K, AC = $$\sqrt 2 $$ k, then ΔABC is a :
A.	ight isosceles triangle
B.	sosceles triangle
C.	ight-angled triangle
D.	quilateral triangle
Answer» B. sosceles triangle

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