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Question Number 183031 by universe last updated on 18/Dec/22

Answered by aleks041103 last updated on 23/Dec/22

$$(x,y)\in T\left(G\right)$$$$\Rightarrow \frac{\pi }{2}s(1-t)=x$$$$\frac{\pi }{2}(1-s)=y\Rightarrow s=1-\frac{2y}{\pi }\Rightarrow x=(\frac{\pi }{2}-y)(1-t)$$$$0<s<1\Rightarrow 0<y<\pi /2$$$$\Rightarrow 0<x<\frac{\pi }{2}-y$$$$\Rightarrow T\left(G\right)isthetrianglex,y>0andx+y<\pi /2$$$$\Rightarrow Area=\frac{1}{2}\left(\frac{\pi }{2}\right)\left(\frac{\pi }{2}\right)=\frac{{\pi }^2}{8}=Area\left(T\left(G\right)\right)$$

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