Question

Please help with geometry

Answer 1

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:
$A=A_O-A_{sg}\to A=144\pi-(24\pi-36\sqrt3)=144\pi-24\pi+36\sqrt3\\\\=\boxed{A=(120\pi+36\sqrt3)m^2}$