We use Co-ordinate Geometry to solve this Problem.
Without ant loss of generality, we may assume that the centre of the
circle is the origin.
A point P, pi/4P,π4 radian on a circle of radius rr have co-ordinates
P(rcos(pi/4), rsin(pi/4))=P(r/sqrt2,r/sqrt2)P(rcos(π4),rsin(π4))=P(r√2,r√2).
Similarly, a point Q, pi/3Q,π3 radians on the circle is Q(r/2,(rsqrt3)/2)Q(r2,r√32).
Given that PQ=3 rArr PQ^2=9PQ=3⇒PQ2=9
rArr (r/sqrt2-r/2)^2+(r/sqrt2-(rsqrt3)/2)^2=9⇒(r√2−r2)2+(r√2−r√32)2=9.
rArr ((rsqrt2)/2-r/2)^2+((rsqrt2)/2-(rsqrt3)/2)^2=9⇒(r√22−r2)2+(r√22−r√32)2=9.
rArr{r/2(sqrt2-1)}^2+{r/2(sqrt2-sqrt3)}^2=9⇒{r2(√2−1)}2+{r2(√2−√3)}2=9.
rArr r^2/4{(2-2sqrt2+1)+(2-2sqrt6+3)}=9⇒r24{(2−2√2+1)+(2−2√6+3)}=9
rArr r^2/4(8-2sqrt2-2sqrt6)=9⇒r24(8−2√2−2√6)=9.
rArr r^2=36/(8-2sqrt2-2sqrt6)⇒r2=368−2√2−2√6
Taking, sqrt2~=1.414, sqrt6~=2.449√2≅1.414,√6≅2.449, we have,
r^2=36/(8-2.828-4.898)=36/0.274=131.39r2=368−2.828−4.898=360.274=131.39, so,
the area of the circle = pir^2=3.14*131.39=412.56=πr2=3.14⋅131.39=412.56