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a and b are real numbers between 0 and 1, A(a,1) ,B(1,b) and C(0,0) are the vertices of a equilateral triangle ABC, then its area is (a) (7√3+12)/4 (b) (7√3-12)/4 (c) √3/4 (d) None of these....Explain pls.
The answer to is question is (d)(none of these)
Solution:
Reason for contradiction of option (a), the maximum area that a triangle can enclose where 0<a,b<1 is <1 (definitely its area will be smaller than a square of side 1 and coordinates (0,0),(0,1),(1,0),(1,1) ).
However (7root(3)+12)/4 is >1 so it is rejected.
Reason for contradiction of option (b), If such a equilateral triangle ever exists with such coordinates its sides should definitely be >1 because the distance between (0,0) and (a,1) where 0<a<1 is always >1.(Use distance formula)
In such a case the area would be > √3(1)2/4 which means area > √3/4 but (7root(3)-12)/4 is less than √3/4
Reason for contradiction of option (c),
For the area to be √3/4 the sides must be 1, for this a,b can be = 0 but still it will form a right angled triangle with third side root(2) which will not be a equilateral triangle.
So option (c) is incorrect
So the only option left is (d)
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