\(? = {S_{^ \circ }} = \,\pi {r^2}\,\,\,\,\)

\(\left( 1 \right)\,\,\,\,\frac{{{S_{\,{\text{blue}}}} + {S_{\Delta \,ABC}}}}{{{S_{\Delta \,ABC}}}}\,\,\, = \,\,\,\frac{{\frac{1}{2}\pi {r^2}}}{{{S_{\Delta \,ABD}}}}\,\,\,\mathop = \limits^{{\text{post}}\,\,{\text{above!}}} \,\,\,\frac{{\pi {r^2}}}{{4\,\,\,{S_{\Delta \,ABC}}}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{S_{^ \circ }} = \pi {r^2}\, = 4\,\,\,\left( {{S_{\,{\text{blue}}}} + {S_{\Delta \,ABC}}} \right) = 4\,\,\,\left( {{S_{\,{\text{sector}}\,\,ACB}}} \right)\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\theta \,\,{\text{ = }}\,\,{\text{9}}{{\text{0}}^ \circ }\)

Note that statement (1) is equivalent to the geometric configuration in which triangle ABD is right and isosceles, but the radius is "free" , that´s why the bifurcation is trivial:

> Take r = 1, so that ? = pi

> Take r = 2, so that ? = 4*pi

\(\left( 2 \right)\,\,\,\,\frac{{2 + \sqrt 2 }}{6} = \frac{{{\text{peri}}{{\text{m}}_{\Delta \,ABC}}}}{{{S_{\Delta \,ABC}}}}\,\)

In the figure shown, we present two possible geometric configurations, in which we have:

(2a) Triangle ABD right and Isosceles (case explored in (1) above) :

\(\frac{{2 + \sqrt 2 }}{6} = \frac{{{\text{peri}}{{\text{m}}_{\Delta \,ABC}}}}{{{S_{\Delta \,ABC}}}}\, = \frac{{2r + r\sqrt 2 }}{{\frac{1}{2}\left( {r \cdot r} \right)}}\,\,\,\, \Rightarrow \,\,\, \ldots \,\,\, \Rightarrow \,\,\,r = 12\,\,\,\left( {viable!} \right)\,\,\,\, \Rightarrow \,\,\,\,?\,\,\, = \,\,144\pi\)

(2b) Triangle ADC is equilateral and, consequently, triangle ABD is a 30-60-90 triangle:

\(\frac{{2 + \sqrt 2 }}{6} = \frac{{{\text{peri}}{{\text{m}}_{\Delta \,ABC}}}}{{{S_{\Delta \,ABC}}}}\, = \frac{{2r + r\sqrt 3 }}{{\frac{1}{2}\left[ {r \cdot \frac{{r\sqrt 3 }}{2}} \right]}}\,\,\,\,\, \Rightarrow \,\,\, \ldots \,\,\, \Rightarrow \,\,\,r > 0,\,\,\,r \ne 12\,\,\,\left( {viable!} \right)\,\,\,\, \Rightarrow \,\,\,\,?\,\,\, \ne \,\,\,144\pi\)

(1+2) Now we know we necessarily have the first case in the bifurcation shown in statement (2), hence (1+2) is SUFFICIENT.

This solution follows the notations and rationale taught in the GMATH method.

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