Harry Potter's Magical Geometry!

Area of $\Delta ABC = \dfrac{\sqrt {3}}{4} \times 1^2 =\dfrac{\sqrt {3}}{4}$

Semi-perimeter $s = \dfrac{1+1+1}{2} = \dfrac{3}{2}$ , Let $r$ be the inradius.

Area of $\Delta ABC = r.s \Rightarrow \dfrac{\sqrt {3}}{4}= \dfrac{3r}{2} \Rightarrow r = \dfrac{\sqrt {3}}{6}$

Area of in-circle = $\pi r^2 = \dfrac{22}{7} \times \left(\dfrac{\sqrt {3}}{6}\right)^2 = \dfrac{22}{7} \times \dfrac{3}{36} = \dfrac{11}{42}$

$\text{Area of shaded region} = \dfrac{\text{Area of triangle ABC - Area of in-circle}}{2}$

$\text{Area of shaded region} = \dfrac{\dfrac{\sqrt {3}}{4} - \dfrac{11}{42}}{2} = \dfrac{21\sqrt{3} - 22}{168}$

$\Rightarrow a+b+c+d = 21 + 3 + 22 + 168 = \huge\color{#D61F06}{\boxed{214}}$

Area of $\triangle ABC=\frac{\sqrt{3}}{4}$

The we have that height since $A$ also is a median and a bisector so radii of the circle will be $\frac{\text{height}}{3}=\frac{\frac{\sqrt{3}}{2}}{3}= \frac{\sqrt{3}}{6}$

Area of the circle is $(\frac{\sqrt{3}}{6})^2\times \frac{22}{7}=\frac{11}{42}$

Now shaded area is $\frac{\triangle ABC-circle}{2}=\frac{\frac{\sqrt{3}}{4}-\frac{11}{42}}{2}=\frac{21\sqrt{3}-22}{168}$

$\therefore a+b+c+d=21+3+22+168=\boxed{214}$

$In ~ equvilateral~ \Delta~, Area ~\Delta=\dfrac{\sqrt3} 4, ~~ Inradius,~ r= \frac 12~~circumradius~R\\ \therefore~~r=\frac 12*\dfrac{a*b*c}{4*Area}=\frac 12*\dfrac{1*1*1}{4*\frac{\sqrt3} 4} ~~ =\dfrac 1{2*\sqrt3}~~\implies~Area~of~circle=\dfrac {22} 7 * \dfrac 1 {12}=\dfrac{22}{7*12} \\ \therefore~~shaded~~ area=\frac 12*\left \{\dfrac{\sqrt3} 4- \dfrac{22}{84} \right \}= \dfrac{21\sqrt3 - 22}{168}.\\ a+b+c+d=21+3+22+168= ~~\color{#D61F06}{214}$

Since the circle is inscribed in an equilateral triangle, the altitude is the perpendicular bisector that goes through the center of the circle. Connect edge $B$ to the center of the circle, call it $O$

Let the perpendicular bisector that starts at $A$ intersect as $D$ on $BC$

$OD$ is the radius

$BO$ is a perpendicular bisector that goes through the center of the circle. Thus, $\angle{OBC}=30$ , $\angle{DOB}=60$ and $\angle{ODB}=90$

Using the properties of 30-60-90 triangle, $OD=\frac { 1 }{ 2\sqrt { 3 } }$

Area of this semicircle is ${ \frac { \pi (\frac { 1 }{ 2\sqrt { 3 } } ) }{ 2 } }^{ 2 }$ which is $\frac{\pi}{24}$

Area of an equilateral triangle is $\frac { \sqrt { 3 } }{ 4 } { a }^{ 2 }$

Substitute $a=1$ and divide by $2$ , because it's half of the triangle, we get $\frac { \sqrt { 3 } }{ 8 }$

Subtract the area of the half triangle and the semicircle: $\frac { \sqrt { 3 } }{ 8 } -\frac { \pi }{ 24 }$

Which simplifies to $\frac { 21\sqrt { 3 } -22 }{ 168 }$

Thus, our answer is $a+b+c+d=21+3+22+168=\boxed{214}$

length of the median = (1/2)(square root of 3)

length of the radius = (length of the median)/3 = (1/6)(square root of 3)

area of the red region =

(1/2)(1/2) (1/2)(square root of 3) - (1/2)(22/7) [(1/6)(square root of 3)]^2 =

[(21)(square root of 3) - 22]/168

a = 21 , b = 3 , c = 22 and d = 168

Then

a + b + c + d = 214