Latest self Aptitude Question SOLUTION: 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

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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

naina (11 years ago)

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Jasmine Khan  (11 years ago)

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

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naina  (11 years ago)

ans is 7pi/12-(root(3)+1)/2 but approach is not known someone pls explain!!

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Jasmine Khan  (11 years ago)

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

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