• Placement Papers
• /
• Elitmus
• /
• Category

Elitmus Question Papers:

(#M40151543) ELITMUS Category question Keep an EYE

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

Asked In Elitmus rohit sharma (10 years ago)

Unsolved Read Solution (7)

Is this Puzzle helpful?   (6)   (1) Submit Your Solution

Read Solution (7) :

rohit sharma  (10 years ago)

btw ans given is : R^2 *( 2*PI/3 - sqrt(3)/2 )

any other approach ?

Is this solution Helpfull? Yes (5) | No   

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 )

Is this solution Helpfull? Yes (4) | No   

siddharth prakash singh  (10 years ago)

when we draw the required figure, 2 circles are intersecting forming a ellipse as a common area. so to find the common area of required figure we have to find end to end length of ellipse.
also area of ellipse= pi*a*b
from fig we find one of the end to end length = R
so a= R/2
now draw a line from center of any circle which touches the intersecting point of two circles.
from fiq we find that length of this line = R
also we know that if two circles intersect in two points then the line through the centres is perpendicular bisector of the common chord.
so by using pythagorus theorem we can find length of b.
b^2= R^2 - (R^2/4)
b=sqrt3R/2
so the required area= pi*a*b
Pi*R/2*sqrt3R/2
=(sqrt3*pi*R^2)/4

Is this solution Helpfull? Yes (2) | No (5)  

Karthik MD  (10 years ago)

pi* r*r/4 where r is the diameter of the common area formed which is also a ciecle

Is this solution Helpfull? Yes | No (2)  

rohit sharma  (10 years ago)

@SIDDHARTH PRAKASH SINGH
i think its not an ellipse
if u cn prove its ellipse then yr solution is right

Is this solution Helpfull? Yes | No   

siddharth prakash singh  (10 years ago)

@Rohit..
If that is not ellipse, then the length of a and b should come same(in case of circle, a=b=r) but from figure using the method what I have used u can get the different values of a and b.

Alternatively u can just construct two circles of given conditions using compass u can find that resulting fig is ellipse.

Is this solution Helpfull? Yes | No   

Amit Kumar  (10 years ago)

When we draw the image, You can find the solution :

2*(area of sector - area of triangle) of any one circle

Triangle having side R, R and sqrt2R.

Is this solution Helpfull? Yes | No   

Please Login to Read Solution.