How Much is Green?

solution

Let $P$ be the intersection point between segments $BC,\,$ and $BC;$ $M$ is the circumcenter of triangle $\triangle (ABC),\,$ right angled in $B, \,\angle{ABC}=\angle{PMB}=90°,\,$ thus: $|BM| = |AM| = |MC| = \frac {|AC|} {2}=5;\,$ $A_S(BDEM)=5^2=25;$

$\triangle{ABM}$ is isosceles, so $\angle{CAB}=\angle{MBA}$ and $\angle{PBM}=90°-\angle{CAB}=\angle{ACB}\implies\triangle (ABC) \sim \triangle (PMB),\,$ thus:

$\frac {|MP|} {|BM|}=\frac {|BM|} {|BC|};$ $|MP|=\frac {|AB|BM|}{|BC|}=\frac {6 \times 5}{8} = \frac {15} 4;\, \beta=\angle{ACB}$

$A_T(BMP)=\frac 1 {2}|MP||BM|=\frac {75} 8;\,\sin\beta=\frac {|AB}{|AC|} = \frac 3 5;$

$A_ Q(BDEP)=A_S(BDEM)-A_T(BMP)=25-\frac {75} 8=\frac {125} 8;$

$A_T(BMC)={\frac 1 2}|BM||BC|sin\beta={\frac 1 2}\times 5\times 8\times\frac 3 5=12;$

$A_T(MCP)=A_T(BMC)-A_T(BMP)=12-\frac{75} 8=\frac{21} 8;$

$A_ {GREEN}=A_Q(BDEP)+A_T(MCP)=\frac{125} 8+\frac{21} 8 = \boxed{\frac{73} 4}=\boxed{18.25}$

Let the length of segment $\overline{BM}$ be $x$ . By Stewart's Theorem , we have $250+10x^2=500 \rightarrow 10x^2=250 \rightarrow x^2=25 \rightarrow x=5$ . Also, let the point of intersection of $\overline{BC}$ and $\overline{ME}$ be $O$ . Now, note that the area of the green region is $[ BDEM ] + [ BMC ] - 2 [ BMO ]$ = $5^2 + \dfrac{6 \cdot 8}{4} - 2 [ BMO ]$ . Lastly, note that $\Delta BMO \sim \Delta ABC$ . Thus, we get that the area of $\Delta BMO$ is $\dfrac{5 \cdot \left( \dfrac{3}{4} \cdot 5 \right)}{2} = \dfrac{75}{8}$ . Thus, our answer is $25+12-2(\dfrac{75}{8}) = 37-\dfrac{75}{4} = 37 - 18.75 = \boxed{18.25}$