Find the area of the grey part

$\triangle ABC\text{ is 30-60-90 } \Rightarrow BC=\sqrt { 3 } \text{ and } AC=2\\ AD\text{ is the angle bisector of } \angle BAC\\ \text{ so } \angle DAC=30º=\angle DCA\Rightarrow AD=DC\\ Now\\ BC=BD+DC=\sqrt { 3 } \rightarrow 1\\ \triangle ABD \text{ is right angle triangle} \\ { \left( AB \right) }^{ 2 }+{ \left( BD \right) }^{ 2 }={ \left( DC \right) }^{ 2 }\qquad //AD=DC\\ 1+{ \left( BD \right) }^{ 2 }={ \left( DC \right) }^{ 2 }\rightarrow 2\\ 1\& 2\Rightarrow BD=\frac { 1 }{ \sqrt { 3 } } and\quad DC=\frac { 2 }{ \sqrt { 3 } }$

( $\text{the area of }\triangle)=\frac{1}{2} \times r \times (\text{the triangle's perimeter})$

$\text{so }r_{ \triangle ABD }=\frac { 2Area }{ perimeter } =\frac { \frac { 1 }{ \sqrt { 3 } } }{ 1+\frac { 1 }{ \sqrt { 3 } } +\frac { 2 }{ \sqrt { 3 } } } =\frac { 1 }{ 3+\sqrt { 3 } } \\ \text{similarly } r_{ \triangle ADC }\frac { 2Area }{ perimeter } =\frac { \frac { 2 }{ \sqrt { 3 } } }{ 2+\frac { 2 }{ \sqrt { 3 } } +\frac { 2 }{ \sqrt { 3 } } } =\frac { 1 }{ 2+\sqrt { 3 } } \\ \text{Area of the grey=Area of } \triangle ABC - \text{(Area of the circles)}\\ A=\frac { \sqrt { 3 } }{ 2 } -\pi \left( r^{ 2 }_{ \triangle ADC }+r^{ 2 }_{ \triangle ADC } \right) \approx 0.5$

This question was very entertaining! The most important thing is to work out the radii of the two circles, which can be achieved by the use of trig and some simple knowledge of right angles. This lead us to the two equations: Tan(30)=r/(root 3 over 3)-r Where r is the radius of the smaller circle; And sin(60)=R/(root 3 over 3)-R Which then could be used to find the area of white bits and obtaining the shaded region as 0.5001713347.