Latest Elitmus Aptitude Question SOLUTION: Two circles of radii R passes through centre of each-other, find the common area to both of the circles in terms of R.

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(#M40151593) ELITMUS Numerical Ability question Keep an EYE

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

Asked In Elitmus siddharth prakash singh (10 years ago)

Unsolved Read Solution (8)

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vipul gera  (10 years ago)

2*r^2(pi/3-sqrt3/4)... for sol see the link
http://www.algebra.com/algebra/homework/Circles/Circles.faq.question.477375.html

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siddharth prakash singh  (10 years ago)

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 )

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akanksha  (9 years ago)

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).

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Pushpak  (10 years ago)

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

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MR. X  (10 years ago)

(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.

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vasuki  (10 years ago)

r^2(2pi/3-sqrt3/2)

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akanksha jain  (9 years ago)

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

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tanmay das  (9 years ago)

plz contect www.geomaths .com

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