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

A. Right isosceles triangle

B. Isosceles triangle

C. Right-angled triangle

D. Equilateral triangle

Answer: Option A

Solution(By Examveda Team)

∵ AB = BC = K
⇒ AC = $$\sqrt 2 $$ K
⇒ (AC)² = (AB)² + (BC)²
⇒ ($$\sqrt 2 $$ K)² = K² + K²
⇒ 2K² = 2K²
⇒ Therefore ΔABC will be a right isosceles triangle.