Latest TCS Aptitude Question SOLUTION: Two identical circles intersect so that their centers , and their point of which they intersect , form a square of 1cm. Find the area of the portion which is common to the two circ

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(#M40028169) TCS Numerical Ability question circles Keep an EYE

Two identical circles intersect so that their centers , and their point of which they intersect , form a square of 1cm. Find the area of the portion which is common to the two circles.

Asked In TCS PRIYA NAMDEO (10 years ago)

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mahak saha  (10 years ago)

area of square=1 cm^2
area of one sector=90/360*3.14*1 cm^2 =0.785 cm^2
area of remaining portion=area of sq.-area of one sector=.215cm^2
common area of both the circle= 1-(2*0.215)=0.572cm^2

Is this solution Helpfull? Yes (39) | No (6)  

Adhideb Rit  (9 years ago)

Suppose O and R the centres of the two circles and OR,AB the two diagonals of the square whose area is 1 sq. cm.
Then edges of square is OA=OB=BR=AR=1 cm.Let intersection point of the two diagonals is P.
Then OP=PR=1/2 cm.

Now we know that 2*pi creates pi*r*r sq. unit area of a circle.
Then θ creates (θ*r*r)/2 sq unit area.
here θ is the angle of the square i.e. angle BOA=angle ARB=θ=pi/2. and edge of square r=1 unit.
So area of OARB=((pi/2)*1*1/2)=pi/4 sq. cm.
Then area created by chord BR or AR or OA or OB is= (Area of OARB - area of square)/2.
So all areas created by these four chords(BR,AR,OA and OB)is 4*((Area of OARB - area of square)/2).
so the area of the portion which is common to the two circles is 4*((Area of OARB - area of square)/2) + area of the square.
=4*((pi/4-1)/2) + 1
=pi/2 - 1

Is this solution Helpfull? Yes (15) | No (3)  

Mani kandan  (9 years ago)

Now the center of the circles and the point of intersection of the circles make a square of side of 1 cm.
Now the area of the common portion can be calculated as = 2(Area of the sector of a circle having the angle of 90o - area of the right angled triangle of side 1 cm). Here the right angled triangle is obtained by joining the two intersecting points which is in turn the diagonal of the square.
Now the area of common section is = 2( (90/360)* pi * 1*1 - (1/2)*1*1)
= (pi/2) - 1

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nipun sabharwal  (10 years ago)

area of sector (angle is 90 and radius is 1) = (pie*1^2)/4 = 3.14/4 = .785
area of small triangle sides( 1,1,root2) = .499 (herons formula)
difference = .286
total area = .286*2 = .572 ans

Is this solution Helpfull? Yes (10) | No (4)  

Olivia Pramanik  (10 years ago)

radius=side of the square=r=1
area intersected by the square on each circle=(pi*(r^2)*a)/360
=(pi*(r^2)*90)/360 ..(a=90 degree)
area intersected by the square on both circle=2*((pi*(r^2)*90)/360)=pi/2

suppose,area of common portion of two circle=x
so,(pi/2)-x=1
=>x=(pi/2)-1
=>x=.57 cm^2

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mahak saha  (10 years ago)

sorry answer is 0.57 cm^2

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Guru Pranav  (10 years ago)

the area of the required sector is 0.570. Ans

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Kuntal Banerjee  (9 years ago)

suppose center of two circle is o and p.and two circle intersect points are a and b.so area of square oapb is=1 cm^2.Now area of triangle oab=triangle bpa=1/2*oapb.So area of triangle oab=1/2 cm^2.And area of arc oab=1/4*area of circle with center o.Area of circle with center o is=pi*1*1=pi.so area of arc oab=pi/4.so area between arc and triangle is=(pi/4)-(1/2).So another circle calculation is same.so total intersected area=2*(pi/4-1/2)=(pi/2)-1

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Pankaj  (7 years ago)

To start with,if ABCD is a square then the angle ∠BAD and ∠BCD are 90 degree. So the Arc ABD and BCD is 1/4th of the circle.
Area of shaded region BD

=> Area of the arc ABD - Area of triangle ABD + Area of the arc CBD - Area of triangle CBD

Also, since the square side is 1cm, the radius of circle is also 1cm.

Hence the Area of each circle = π * 1 * 1 = π

Area of arc ABD = π / 4

Area of arc CBD = π / 4

Area of Triangle ABD = ( 1/2 ) * Area of Square ABCD = ( 1/2 ) * 1 * 1 = 1/2

Area of Triangle CBD = ( 1/2 ) * Area of Square ABCD = ( 1/2 ) * 1 * 1 = 1/2

Putting all these values in EQ1 we get

=> Area of the arc ABD - Area of triangle ABD + Area of the arc CBD - Area of triangle CBD

=> π / 4 - 1/2 + π / 4 - 1/2

=> π / 4 + π / 4 -1/2 - 1/2

=> π / 2 - 1

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prakher bhargava  (10 years ago)

area of segment =area of sector- area of triangle =((theta/360)*pie *r^2)-(1/2 r^2 sin theta )=.214 for 1 part of circle ..since there are two common circle and a square therefor 2*.214=.428... @mahak why are u subtracting this from 1...can u pls answer...because by apply formula u get the segment.directley

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amey misal  (10 years ago)

the required portion will be 2(3.14*1*1-0.5*1*1)....
subtract the area of triangle formed in the circle from area of circle..do for boh the circles and add...

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kumar  (10 years ago)

@NIPUN: WE CAN USE 1/2*BASE^2 FOR FINDING AREA ,BECAUSE TWO SIDE WILL BE 1CM EACH AND ANGLE BETWEEN THEM WILL BE 90.

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B.Nikitha  (8 years ago)

when two circles are intersected with each other and form a square with side1cm the area of common portion in both the circles is 2(area of sector in circle-(1/2)area of aquare)
so area of sector=(90/360)*pi*r*r
2[((1/4)pi*1*1)-(1/2)1]=(pi/2)-1
so the ans is pi/2-1

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sunayan das  (7 years ago)

So, the side of the square is 1, which means the radius of each circle is also 1.
Area of squares = 1
Area of two circles not included in the square = 3(pi)/2
Area of oval overlap region of circles = [(pi)/2] - 1
Area of entire diagram = 3(pi)/2 + 1
suppose center of two circle is o and p.and two circle intersect points are a and b.so area of square oapb is=1 cm^2.Now area of triangle oab=triangle bpa=1/2*oapb.So area of triangle oab=1/2 cm^2.And area of arc oab=1/4*area of circle with center o.Area of circle with center o is=pi*1*1=pi.so area of arc oab=pi/4.so area between arc and triangle is=(pi/4)-(1/2).So another circle calculation is same.so total intersected area=2*(pi/4-1/2)=(pi/2)-1

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yelapu girish ram pradeep  (4 years ago)

let us assume two circles intersecting each other at two points.
now join the lines from center to intersecting points .total four lines form square of side 1cm.
if we observe the figure it is the joining of two cones with same lateral surface area.
for calculating it height of cone is half of the diagonal value.
diagonal=square root of 1+1
that equals square root of 2
so height of cone is (square root of 2)/2
lateral surface area of cone is portion of intersection so formula is pi*r*square root of sum of the square of height and radius.
in this problem height=radius
therefore,
intersecting portion equals to pi*0.707*1=2.22

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