In the figure, θ _1+θ _2= π 2 and √3( BE)=4( AB). If the area of Δ CAB is 2√3-3 unit^2, when θ _2 θ _1 is the largest, then the perimeter (in unit)…

Given,

And ${\theta }_{1}+{\theta }_{2}=\frac{\pi }{2}$θ1​+θ2​=2π​

Now, Let $\mathrm{AB}=x,\mathrm{BD}=y$AB=x,BD=y

Also given, $\sqrt{3}\mathrm{BE}=4\mathrm{AB}$3​BE=4AB

$$\Rightarrow \sqrt{3}(y+\mathrm{DE})=4x$$⇒3​(y+DE)=4x

$$\Rightarrow \mathrm{DE}=\frac{4x}{\sqrt{3}}-y$$⇒DE=3​4x​−y

Now given area of triangle $\Delta \mathrm{CAB}=2\sqrt{3}-3$ΔCAB=23​−3

$$\Rightarrow \frac{1}{2}xy=2\sqrt{3}-3$$⇒21​xy=23​−3

$$\Rightarrow y=\frac{4\sqrt{3}-6}{x}$$⇒y=x43​−6​

Now finding,

$${tan\theta }_{2}=\frac{(\frac{4x}{\sqrt{3}}-y)}{x}=\frac{4}{\sqrt{3}}-\frac{(4\sqrt{3}-6)}{{x}^{2}}$$tanθ2​=x(3​4x​−y)​=3​4​−x2(43​−6)​

$${tan\theta }_{1}=\frac{y}{x}=\frac{4\sqrt{3}-6}{{x}^{2}}$$tanθ1​=xy​=x243​−6​

Now taking $\mathrm{tan}$tan both of ${\theta }_{1}+{\theta }_{2}=\frac{\pi }{2}$θ1​+θ2​=2π​ we get,

$${tan\theta }_{1}\cdot {tan\theta }_{2}=1$$tanθ1​⋅tanθ2​=1

$$\Rightarrow (\frac{4}{\sqrt{3}}-\frac{4\sqrt{3}-6}{{x}^{2}})\cdot (\frac{4\sqrt{3}-6}{{x}^{2}})=1$$⇒(3​4​−x243​−6​)⋅(x243​−6​)=1

$$\Rightarrow \frac{4\sqrt{3}-6}{{x}^{2}}=\sqrt{3}\mathrm{or}\frac{1}{\sqrt{3}}$$⇒x243​−6​=3​or3​1​

So, $\frac{{\theta }_{2}}{{\theta }_{1}}$θ1​θ2​​ is maximum when $\frac{4\sqrt{3}-6}{{x}^{2}}=\frac{1}{\sqrt{3}}$x243​−6​=3​1​

$$\Rightarrow {x}^{2}=\sqrt{3}(4\sqrt{3}-6)$$⇒x2=3​(43​−6)

$$\Rightarrow {x}^{2}=12-6\sqrt{3}$$⇒x2=12−63​

$$\Rightarrow {x}^{2}={(3-\sqrt{3})}^{2}$$⇒x2=(3−3​)2

$\Rightarrow x=3-\sqrt{3}$⇒x=3−3​ and ${\theta }_{2}=60^{\circ}$θ2​=60∘

$$\Rightarrow \mathrm{tan}60^{\circ}=\frac{\mathrm{DE}}{\mathrm{CD}}\Rightarrow \mathrm{DE}=x\cdot \sqrt{3}=3\sqrt{3}-3$$⇒tan60∘=CDDE​⇒DE=x⋅3​=33​−3

And $\mathrm{cos}60^{\circ}=\frac{\mathrm{CD}}{\mathrm{CE}}\Rightarrow \mathrm{CE}=2x=6-2\sqrt{3}$cos60∘=CECD​⇒CE=2x=6−23​

So, the Perimeter of $\Delta \mathrm{CED}=\mathrm{CD}+\mathrm{DE}+\mathrm{CE}=6$ΔCED=CD+DE+CE=6