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.

`(60^o)/(360^o)=(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=(1)/(6)\cdot\pi\cdot12^2=(1)/(6)\pi\cdot144=24\pi\ m^2`
The formula of an area of the equilateral triangle:
`A_\Delta=(a^2\sqrt3)/(4)`

`A_\Delta=(12^2\sqrt3)/(4)=(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}`

Answer 2

`\pi . {12}^(2) . (300)/(360) + {12}^(2) ( √(3) )/(4) \\ {12}^(2) ( (5.\pi)/(6) + ( √(3) )/(4) ) \\( 120\pi + 36 √(3) ) {cm}^(2)`