2 circles intersect at pt ab arc made by smaller circle is 90 degree and that of larger circle is 60 degree find area of intersection

• by mistake i wrote + sign, it is - sign.
  let us draw the diagram first.
  c and d be the center of the circles and a & b are the points of intersection.
  given angle acb=90 degree=pi/2 radian and angle adb=60 degree=pi/3 radian
  therefore angle cab=cba=45 degree and angle dab=dba=60 degree
  by using 45-45-90 and 30-60-90 triangle we found that radius of the first circle having center c is r1=1 and the radius of the circle having center d is r0=root2.
  area=(1/2)*[angle adb*r0^2-r0^2*sin(adb)+angle acb*r1^2-r1^2*sin(acb)]
  (remember all the angles must be written in radian only.)
  area=(1/2)*[(pi/3)*2-(root3/2)*2+*(pi/2)-1]
  area=pi/3-root3/2+pi/4-1/2
  area=7pi/12-(root3+1)/2
  hence the answer.
  if you are still not getting it contact me at jasmine.khan@seedinfotech.com
• 11 years agoHelpfull: Yes(1) No(0)
• ans is 7pi/12-(root(3)+1)/2 but approach is not known someone pls explain!!
• 11 years agoHelpfull: Yes(0) No(0)
• let us draw the diagram first.
  c and d be the center of the circles and a & b are the points of intersection.
  given angle acb=90 degree=pi/2 radian and angle adb=60 degree=pi/3 radian
  therefore angle cab=cba=45 degree and angle dab=dba=60 degree
  by using 45-45-90 and 30-60-90 triangle we found that radius of the first circle having center c is r1=1 and the radius of the circle having center d is r0=root2.
  area=(1/2)*[angle adb*r0^2-r0^2*sin(adb)+angle acb*r1^2-r1^2*sin(acb)]
  (remember all the angles must be written in radian only.)
  area=(1/2)*[(pi/3)*2-(root3/2)*2+*(pi/2)-1]
  area=pi/3-root3/2+pi/4-1/2
  area=7pi/12+(root3+1)/2
  hence the answer.
  if you are still not getting it contact me at jasmine.khan@seedinfotech.com
• 11 years agoHelpfull: Yes(0) No(0)