# Can I hatch this region in any way?

Can I hatch this region in any way?

Graphics[{
Circle[{0,0},10,{ArcCos[5/10],ArcCos[5Sqrt/10]}],
Line[{{5Sqrt,5},{5Sqrt,-5}}],
Circle[{0,0},10,{-ArcCos[5/10],-ArcCos[5Sqrt/10]}],
Line[{{5,-5Sqrt},{5,5Sqrt}}]
},
Axes->True,
AxesOrigin->{0,0}
] EDIT

I want to define a region of a circle, because I want to determine the area of this region... • For what purpose? – Alex Trounev Jan 3 '19 at 19:10

reg = ImplicitRegion[x^2 + y^2 <= 100 && 5 <= x <= 5 Sqrt, {x, y}];
N[Area[reg]]


52.3599

Show[Graphics[{Gray, Circle[{0, 0}, 10]}, Axes -> True],
RegionPlot[reg, MeshFunctions -> {# + #2 &, # - #2 &},
Mesh -> {50, 50}, MeshShading -> None, PlotStyle -> None,
BoundaryStyle -> Red]] RegionPlot[x^2 + y^2 <= 100 && 5 <= x <= 5 Sqrt,{x, -10, 10}, {y, -10, 10},
MeshFunctions -> {# + #2 &, # - #2 &}, Mesh -> {50, 50},
MeshShading -> None, PlotStyle -> None, BoundaryStyle -> Red,
PlotPoints -> 90, Axes -> True, Epilog -> {Gray, Scale[Circle[], 10]},
Frame -> False]


same picture

poly = MeshPrimitives[
BoundaryDiscretizeRegion[
RegionIntersection[
Disk[],
HalfPlane[{{5/10, 0}, {5/10, 1}}, {1, 0}],
HalfPlane[{{5 Sqrt/10, 0}, {5 Sqrt/10, 1}}, {-1, 0}]
],
MaxCellMeasure -> {1 -> 0.001}
],
2
][];
Area[poly]


0.523599

Graphics[{Circle[], Gray, EdgeForm[{Thick, Black}], poly}] Graphics[{Red, Opacity@0.7, Disk[{0, 0}, 10], Opacity@1, Blue, Thick,
Circle[{0, 0}, 10, {π/6, π/3}],
Circle[{0, 0}, 10, {-(π/6), -(π/3)}], Green, Opacity@0.6,
Rectangle[{10 Cos[π/3], -10 Sin[π/3]}, {10 Cos[π/6],
10 Sin[π/3]}]}, Axes -> True, AxesOrigin -> {0, 0}] Therefore we can use the following.

reg1 = Disk[{0, 0}, 10];
reg2 = Rectangle[{10 Cos[π/3], -10 Sin[π/3]}, {10 Cos[π/ 6], 10 Sin[π/3]}];
reg=RegionIntersection[reg1, reg2];
Area@reg


$$\frac{50 \pi }{3}$$

Show[Graphics[{Circle[{0, 0}, 10]}, Axes -> True],
Region[reg, BaseStyle -> {LightBlue, EdgeForm[{Red, Thick}]}]] You can also choose reg2 as

reg2 = Rectangle[{5, -5 Sqrt}, {5 Sqrt, 5 Sqrt}];


Or

reg2 = Rectangle[{5, -10}, {5 Sqrt, 10}];