Two circles of radii R passes through centre of each-other, find the common area to both of the circles in terms of R.

• 2*r^2(pi/3-sqrt3/4)... for sol see the link
  http://www.algebra.com/algebra/homework/Circles/Circles.faq.question.477375.html
• 9 years agoHelpfull: Yes(24) No(4)
• let two centres = O and O' and two intersection points as A and B.
  join O,A ,O' AND B. A rhombus of side R will be formed.
  Now, area of req region= (area of two sectors forming with center of 2 circles - area of two triangles having two sides as radius and one side as AB)
  join two centres of two circles O and O'.
  this will divide the rhombus OAO'B into two equilateral triangles(OAO' and OBO') of side R.
  so /_A=/_AOO'=/_AO'O=60
  and /_B=/_BOO'=/_BO'O=60
  hence /_AOB+/_AO'B=240
  => /_AOB=/_AO'B=120
  so area of 2 sectors= 2*pi*R^2*120/360 =2*pi*R^2/3
  since area of triangle=1/2*product of two sides* sine of anle b/w them.
  so sum of area of triangles AOB AND AO'B= 2*1/2*R^2*sine 120 =sqrt(3)*R^2/2
  Hence area of req region= =2*pi*R^2/3 - sqrt(3)*R^2/2
  = R^2 *( 2*pi/3 - sqrt(3)/2 )
• 9 years agoHelpfull: Yes(15) No(3)
• let two centers=o and o' and two intersection points p and q.join o and o'.Draw the radii,We will get a rhombus.angle at the center will be 120.now area of a segment of circle is (theta/360*pi r^2-1/2 r^2 sin theta).as the area to be calculated is the area of the segments of both the circle so required area will b 2*(120/360*pi r^2-1/2 r^2 sin 120) i.e r^2(2/3 pi-root 3/2).
• 9 years agoHelpfull: Yes(4) No(0)
• when two circle passes through centre of each other then common area of both of circle
  = 2* area of segment
  =2*{theta/360 *pi r^2 - 1/2 r^2 sin theta} here theta =120
  =1/3 pi r^2- root (3)/4 r^2
  =r^2{pi/3-root 3/4} this is correct ans
• 9 years agoHelpfull: Yes(2) No(1)
• (pi R^2)/2 ans
  circumference of one circle is touching center of other
  (if touching center then the other circle is half covered) .
  
  assumption correctify me if wrong.
  
• 9 years agoHelpfull: Yes(1) No(6)
• r^2(2pi/3-sqrt3/2)
• 9 years agoHelpfull: Yes(1) No(0)
• let two centres = O and O' and two intersection points as A and B.
  join O,A ,O' AND B. A rhombus of side R will be formed.
  D will be the midpoint of O and O' so Triangle OAD will be right angled triangle with sides OA=R OD=R/2 so using pythagoras theorem OA^2=OD^2+AD^2 i.e. AD will be R*sqrt(3)/2..and thus AB will be R*sqrt(3) now as we know AB and OO' an ellipse is formed area of ellipse will be pi*R/2*(R*sqrt(3)/2)=pi*R^2*sqrt(3)/4
  
• 9 years agoHelpfull: Yes(0) No(0)
• plz contect www.geomaths .com
• 9 years agoHelpfull: Yes(0) No(0)