In the figure, θ1 + θ2 = \(\frac{π}{2}\) and √3 (BE) = 4 (AB). If the area of ΔCAB is 2√3 - 3 unit2 , when \(\frac{θ_2}{θ_1}\) is the largest, then the perimeter (in unit) of ΔCED is equal to ________.
Correct Answer: 6
Solution and Explanation
Step 1: Define the tangents and angles. Let the tangent be: \[ y = mx \pm \sqrt{19m^2 + 15} \] Now, using the equation \( mx - y \pm \sqrt{19m^2 + 15} = 0 \) to solve for the parallel line from (0, 0): \[ \left| \frac{\sqrt{19m^2 + 15}}{\sqrt{m^2 + 1}} \right| = 4 \]
Step 2: Solve for \( m \). We get the equation: \[ 19m^2 + 15 = 16m^2 + 16 \] Solving this: \[ 3m^2 = 1 \quad \Rightarrow \quad m = \pm \frac{1}{\sqrt{3}} \]
Step 3: Find the angle. The angle with the x-axis is: \[ \theta = \frac{\pi}{6} \] Thus, the required angle is: \[ \frac{\pi}{3} \]
Step 4: Calculate the perimeter. Now, calculate \( x \) from the area equation: \[ x^2 = 12 - 6\sqrt{3} = (3 - \sqrt{3})^2 \] Hence, \( x = 3 - \sqrt{3} \).
Step 5: Final Calculation. The perimeter of \( \triangle CED \) is: \[ \text{Perimeter} = CD + DE + CE = 3\sqrt{3} + (3\sqrt{3}) + (3 - \sqrt{3}) = 6 \] Thus, the perimeter of \( \triangle CED \) is \( \boxed{6} \).
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