Question
In ΔABC, AB = BC = K, AC = $$\sqrt 2 $$ k, then ΔABC is a :
Answer: Option A
∵ 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.
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∵ 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.
Was this answer helpful ?
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