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 :
(i) A triangle is isosceles, if its any two sides are equal, so to prove a triangle isosceles, we have to prove as two sides are equal. (ii) To prove a triangle a right angled triangle, we have to prove that the sum of square of any two sides is equal to the square of third side. Let points A (4, 0), B (-1, - 1) and C(3, 5) be the vertices of a \[\Delta ABC\]. \[\therefore AB=\sqrt{(-1-{{4}^{2}}+{{(-1-0)}^{2}}}=\sqrt{25+1}=\sqrt{26}\] \[BC=\sqrt{{{(3+1)}^{2}}+{{(5+1)}^{2}}}\] \[=\sqrt{{{4}^{2}}+{{6}^{2}}}\] \[=\sqrt{16+36}=\sqrt{52}\] and \[CA=\sqrt{{{(4-3)}^{2}}+{{(0-5)}^{2}}}=\sqrt{1+25}=\sqrt{26}\] Now, \[C{{A}^{2}}+A{{B}^{2}}={{(\sqrt{26})}^{2}}+{{(\sqrt{26})}^{2}}\] \[=26+26=52=B{{C}^{2}}\] \[\Rightarrow \] \[C{{A}^{2}}+A{{B}^{2}}=B{{C}^{2}}\] Thus, the triangle is isosceles and right angled triangle.
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