The two circles are symmetric about the diagonal. NC‌1=
2√2
3
=√(
2
3
)2+(
2
3
)2=
2√2
3
The lengths FC‌1, EC‌1 are the radius of the circle which is
2km
3
. The length C1P is the radius of the circle. Because of symmetry C1O=C2O and C1N=C2B 2×(C1N+C1O)=2√2 the length of the diagonal of the square. C1N+C1O=√2 C1O=√2−
2√2
3
=
√2
3
OP=(
2−√2
3
) The diagonal perpendicularly bisects the line GH Hence C1‌OH is 90‌degreesC1H2=C1O2+OH2 OH=
√2
3
Similarly OG=
√2
3
HC‌1,C1G both are equal to
2
3
each. H1G is
2√2
3
.HC‌1‌G is a right angled triangle with angle HC‌1‌G is 90‌degrees Area of the region required is 2×(Area‌of‌segment‌OGH) .'. Area of required region =2(