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In ΔABC, AB = BC = K, AC = √2 K, then ΔABC is a: (SSC CHSL 2015)
- a)Isosceles triangle
- b)Right angled triangle
- c)Equilateral triangle
- d)Right isosceles triangle
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
In ΔABC, AB = BC = K, AC = √2 K, then ΔABC is a: (S...
In ΔABC
AC = 2K
AC2 = 2K2
AC2 = AB2 + BC2
So ΔABC is right angled triangle
So, in ΔABC
So, In triangle ABC, ∠B = 90°; ∠C =45°; ∠A = 45°
Hence, triangle ABC is right isoscles triangle.
AC = 2K
AC2 = 2K2
AC2 = AB2 + BC2
So ΔABC is right angled triangle
So, in ΔABC
So, In triangle ABC, ∠B = 90°; ∠C =45°; ∠A = 45°
Hence, triangle ABC is right isoscles triangle.
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Community Answer
In ΔABC, AB = BC = K, AC = √2 K, then ΔABC is a: (S...
In ΔABC
AC = 2K
AC2 = 2K2
AC2 = AB2 + BC2
So ΔABC is right angled triangle
So, in ΔABC
So, In triangle ABC, ∠B = 90°; ∠C =45°; ∠A = 45°
Hence, triangle ABC is right isoscles triangle.
AC = 2K
AC2 = 2K2
AC2 = AB2 + BC2
So ΔABC is right angled triangle
So, in ΔABC
So, In triangle ABC, ∠B = 90°; ∠C =45°; ∠A = 45°
Hence, triangle ABC is right isoscles triangle.
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In ΔABC, AB = BC = K, AC = √2 K, then ΔABC is a: (SSC CHSL 2015)a)Isosceles triangleb)Right angled trianglec)Equilateral triangled)Right isosceles triangleCorrect answer is option 'D'. Can you explain this answer?
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