In the figure, θ 1 + θ 2 = π 2 and 3 BE = 4 AB . If the area of ∆ CAB is 2 3 - 3 unit 2 , when θ 2 θ 1 is the largest, then the perimeter (in unit) of ∆ CED is equal to

Step-by-step solution

Given, And θ 1 + θ 2 = π 2 Now, Let AB   = x ,   BD = y Also given, 3 BE = 4 AB ⇒    3 y + DE = 4 x ⇒ DE = 4 x 3 - y Now given area of triangle ∆ CAB = 2 3 - 3 ⇒    1 2 x y = 2 3 - 3 ⇒    y = 4 3 - 6 x Now finding, tanθ 2 = 4 x 3 - y x = 4 3 - ( 4 3 - 6 ) x 2 tanθ 1 = y x = 4 3 - 6 x 2 Now taking tan both of θ 1 + θ 2 = π 2 we get, tanθ 1 · tanθ 2 = 1 ⇒ 4 3 - 4 3 - 6 x 2 · 4 3 - 6