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InterviewSolution
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The sides of a triangle are in A.P. and its area is `(3)/(5)` th of an equilateral triangle of the same perimeter. Find the greatest angle of the triangle |
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Answer» Correct Answer - `120^(@)` Let the sides be `x - d, x, x + d`. Then `s = (3x)/(2), (s -a) = (x)/(2) + d`, `(s-b) = (x)/(2), (s-c) = (x)/(2) -d`. Area of triangle `= sqrt((3x)/(2).((x)/(2) + d).(x)/(2).((x)/(2) -d))` `= (x)/(2) sqrt(3((x^(2))/(4) -d^(2))) = (x)/(4) sqrt(3(x^(2) -4d^(2)))` The area of equilateral triangle whose perimeter is `3x " is " (sqrt3x^(2))/(4)` Given, `(3)/(5) xx (sqrt3x^(2))/(4) = (x)/(4) sqrt(3(x^(2) -4d^(2)))` or `(9)/(25) xx (3x^(4))/(16) = (x^(2))/(16) xx 3 (x^(2) -4d^(2))` or `x^(2) - (9x^(2))/(25) = 4d^(2)` or `16 x^(2) = 100d^(2)` or `x = (5)/(2) d` Therefore, the sides of triangle measure `((5d)/(2) -d), (5d)/(2), ((5d)/(2) + d)` or `(3d)/(2), (5d)/(2), (7d)/(2)` Hence, the ratio of sides is `3 : 5 : 7` For the greatest angle, `cos theta = (3^(2) + 5^(2) -7^(2))/(2 xx 3 xx 5) = (9 + 25 - 49)/(30) = (-1)/(2)` `:. theta = 120^(2)` |
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