A) equilateral triangle
B) isosceles triangle
C) scalene triangle
D) right triangle
Correct Answer: B
Solution :
The coordinates of the vertices A, B and C are \[\left( \frac{9}{2},9 \right),\]\[(3,7)\]and \[(6,7)\]respectively. |
Now, \[AB=\sqrt{{{\left( 3-\frac{9}{2} \right)}^{2}}+{{(7-9)}^{2}}}=\sqrt{\frac{9}{4}+4}=\frac{5}{2}=2.5\] units |
\[BC=\sqrt{{{(6-3)}^{2}}+{{(7-7)}^{2}}}=3\]units |
and \[AC=\sqrt{{{\left( \frac{9}{2}-6 \right)}^{2}}+{{(9-7)}^{2}}}=\sqrt{\frac{9}{4}+4}=\frac{5}{2}=25\] units |
As, two sides AB and AC are equal, it is an isosceles triangle. |
Moreover, \[B{{C}^{2}}\ne A{{B}^{2}}+A{{C}^{2}}\] |
\[\therefore \] ABC is not a right triangle. |
So, option [b] is correct, |
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