A) an equilateral triangle
B) an isosceles triangle
C) a right triangle
D) a right angled isosceles triangle
Correct Answer: A
Solution :
| [a] Let ABC be the triangle whose vertices are \[A(2,4),\]\[B(2,6)\]and \[C(2+\sqrt{3},5)\] |
| \[\therefore \,\,\,AB=\sqrt{{{(2-2)}^{2}}+{{(6-4)}^{2}}}=\sqrt{{{0}^{2}}+{{2}^{2}}}=2\] |
| \[BC=\sqrt{{{(2+\sqrt{3}-2)}^{2}}+{{(5-6)}^{2}}}=\sqrt{{{(\sqrt{3})}^{2}}+{{(-1)}^{2}}}=\sqrt{4}=2\]\[AC=\sqrt{{{(2+\sqrt{3}-2)}^{2}}+{{(5-4)}^{2}}}=\sqrt{{{(\sqrt{3})}^{2}}+{{1}^{2}}}=\sqrt{4}=2\]We have, \[AB=BC=AC=2\] |
| Hence, \[\Delta ABC\]is an equilateral triangle . |
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