A) isosceles and right angled
B) isosceles but not right angled
C) right angled but not isosceles
D) neither right angled nor isosceles
Correct Answer: A
Solution :
\[AB=\sqrt{{{(4+1)}^{2}}+{{(0+1)}^{2}}}=\sqrt{26}\] \[BC=\sqrt{{{(3+1)}^{2}}+{{(5+1)}^{2}}}=\sqrt{52}\] \[CA=\sqrt{{{(4-3)}^{2}}+{{(0-5)}^{2}}}=\sqrt{26}\] So, in isosceles triangles side\[AB=CA\] For right angled triangle \[B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}\] So, here \[BC=\sqrt{52}=B{{C}^{2}}=52\] or \[{{(\sqrt{26})}^{2}}={{(\sqrt{26})}^{2}}=52\] So, given vertices is right angled and also is isosceles triangle.You need to login to perform this action.
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