A chord with a length of 3 runs from pi/4 to pi/3 radians on a circle. What is the area of the circle?

1 Answer
Aug 2, 2016

412.56.

Explanation:

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/4 radian on a circle of radius r have co-ordinates

P(rcos(pi/4), rsin(pi/4))=P(r/sqrt2,r/sqrt2).

Similarly, a point Q, pi/3 radians on the circle is Q(r/2,(rsqrt3)/2).

Given that PQ=3 rArr PQ^2=9

rArr (r/sqrt2-r/2)^2+(r/sqrt2-(rsqrt3)/2)^2=9.

rArr ((rsqrt2)/2-r/2)^2+((rsqrt2)/2-(rsqrt3)/2)^2=9.

rArr{r/2(sqrt2-1)}^2+{r/2(sqrt2-sqrt3)}^2=9.

rArr r^2/4{(2-2sqrt2+1)+(2-2sqrt6+3)}=9

rArr r^2/4(8-2sqrt2-2sqrt6)=9.

rArr r^2=36/(8-2sqrt2-2sqrt6)

Taking, sqrt2~=1.414, sqrt6~=2.449, we have,

r^2=36/(8-2.828-4.898)=36/0.274=131.39, so,

the area of the circle = pir^2=3.14*131.39=412.56