A) An acute angled triangle
B) A right angled triangle
C) An isosceles triangle
D) None of these
Correct Answer: A
Solution :
\[AB=\sqrt{{{(2a-2a)}^{2}}+{{(4a-6a)}^{2}}}=2a\] \[BC=\sqrt{{{(\sqrt{3}a)}^{2}}+{{a}^{2}}}=2a\] \[CA=\sqrt{{{(\sqrt{3}a)}^{2}}+{{(-a)}^{2}}}=2a\] Since \[AB=BC=CA,\] hence triangle is equilateral. Therefore, it is an acute angled triangle.
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