The Dog's Span II

The pink,yellow and green sections are the ones the dog can reach to. The area of the red sector is : $A=\frac { 5\quad \times \quad \pi \quad \times \quad { r }^{ 2 } }{ 6 } =\frac { 5\quad \times \quad \pi \times \quad { 12 }^{ 2 } }{ 6 } =120\pi$ Because the triangle is equilateral, the angular sector makes an angle of 360-60=300=360*5/6

Now we have to determine the area formed by the triangle SAC. S is on the cercle of center C and radius 7 so SC=7, S is also on the cercle of center A and radius 7, so SA=7. Therefore SAC is isosceles. Therefore (SZ) is the perpendicular bisector of segment [AC]. Therefore using the Pythagorean theorem : $SZ=\sqrt { { SC }^{ 2 }-{ ZC }^{ 2 } } =\sqrt { { 7 }^{ 2 }-{ 2.5 }^{ 2 } } =\sqrt { 42.75 }$

Thus, $Ayellow=\frac { AC\times SZ }{ 2 } =\frac { 5\times \sqrt { 42.75 } }{ 2 }$ =16.346

Now for the green angular sectors. We have to find the angle $\widehat { EAS }$ . $\widehat { EAC } +\widehat { CAB } =180\\ So,\quad \widehat { EAC } =120$ $\widehat { ZAS } =\cos ^{ -1 }{ (\frac { AZ }{ AS } ) } =\quad \cos ^{ -1 }{ (\frac { 2.5 }{ 7 } ) } \\ Therefore,\\ \widehat { EAS } =120-\cos ^{ -1 }{ (\frac { 2.5 }{ 7 } ) } \simeq 50.925\\ Then,\quad A\quad green\quad =\quad 2\times \frac { \pi \times { 7 }^{ 2 }\times 50.925 }{ 360 } \simeq 13.863\pi \\ Finally\quad \Alpha total=120\pi +13.863\pi +16.346=436.88\\ The\quad answer\quad is\quad 436.$