Question

Please help with geometry

Answer 1

$\pi . {12}^{2} . \dfrac{300}{360} + {12}^{2} \dfrac{ \sqrt{3} }{4} \\ {12}^{2} ( \dfrac{5.\pi}{6} + \dfrac{ \sqrt{3} }{4} ) \\( 120\pi + 36 \sqrt{3} ) {cm}^{2}$

Answer 2

It's a equilateral triangle. The area of the white region is the area of a sector minus the area of a triangle.
$\dfrac{60^o}{360^o}=\dfrac{1}{6}$ 
The sector is 1/6 of the circle, therefore the area of the sector is 1/6 of area of the circle.
 The formula of an area of the circle: $A_O=\pi r^2$ 
 The area of the sector: 
$A_s=\dfrac{1}{6}\cdot\pi\cdot12^2=\dfrac{1}{6}\pi\cdot144=24\pi\ m^2$
 The formula of an area of the equilateral triangle: $A_\Delta=\dfrac{a^2\sqrt3}{4}$
 $A_\Delta=\dfrac{12^2\sqrt3}{4}=\dfrac{144\sqrt3}{4}=36\sqrt3\ m^2$  

The area of the segment: $A_{sg}=A_s-A_\Delta\to A_{sg}=(24\pi-36\sqrt3)m^2$
  The area of the shaded region: 
$A=A_O-A_{sg}\to A=144\pi-(24\pi-36\sqrt3)=144\pi-24\pi+36\sqrt3\\\\=\boxed{A=(120\pi+36\sqrt3)m^2}$