Question 135

In $$\triangle ABC, AB=BC=K, AC=\sqrt2\ K$$, then $$\triangle ABC$$ is a:

Solution

The three sides are not equal, hence it is not an equilateral triangle.

Now, $$(AB)^2+(BC)^2=(k)^2+(k)^2=2k^2$$

Also, $$(AC)^2=(\sqrt2\ k)^2=2k^2$$

$$\because$$ $$(AB)^2+(BC)^2=(AC)^2$$

Thus, $$\triangle$$ ABC is a Right isosceles triangle.

=> Ans - (D)

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