Question

In $△ABC,$ AB=BC=K, AC =$\sqrt{2}k,$ then $△ABC$ is a :

• A. Isosceles triangle
• B. Right-angled triangle
• C. Equilateral triangle
• D. Right isosceles triangle

Answer 1

Answer: (A), (B)

The correct options are
A Isosceles triangle
B Right-angled triangle

$AB=BC=K$

Two sides are equal

$\Rightarrow$ ISosceles triangle

$AC=\sqrt{2}K$

$\Rightarrow$ ${AC}^2=2K^2$

${AB}^2+{BC}^2=K^2+K^2=2K^2$

$\Rightarrow$ ${AC}^2={AB}^2+{BC}^2$

$\Rightarrow$ Right angle triangle also

$\Rightarrow$ Right isosceles triangle