Latest TCS Aptitude Question SOLUTION: two identical circles intersect so that their circles and the point at which they intersect form a square of side 1 cm. The area in sq cm of the portion common to both the circles.

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two identical circles intersect so that their circles and the point at which they intersect form a square of side 1 cm. The area in sq cm of the portion common to both the circles.

Asked In TCS Twinkle (9 years ago)

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RAKESH  (9 years ago)

area of sector = pi*r^2*90/360 = pi/4 = [r=1=side of square]

area of triangle = 1/2 *1*1 = .5

area of common portion = 2*[area of sector - area of triangle]
=2*(pi/4-0.5)
= 1.57 - 1
= .57 cm^2

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Tapas  (9 years ago)

First draw the circles on the paper which will intersect each other and whose centres will be in a line.
Let assume 1st circle's centre is "A" & 2nd's is "C"
now join the intersecting points with the centres. Now the points will be "A(first circle's centre)", "B(first intersecting point)",
"C(2nd centre)", "D(2nd intersecting point)"..
now shade the common region as black(assume)
now we have to find the shaded regions area in sq. m.
According to the question the radious is 1 m means arm of the square is 1 m(i.e AB=BC=CD=DA=1)
now area of the region "ABDA" is (Pi/4*r^2) similarly the area of the region "BCDB" is (Pi/4*r^2)
so the area of the 2 arcs is 2*(Pi/4*r^2)=(Pi/2*r^2)
& the area of the square "ABCD" is 1 sq m.
so the shaded regions area is={area of the 2 arcs("ABDA"+"BCDB")-area of the square("ABCD")}
=>(Pi/2*r^2)-1
=>(Pi/2*1^2)-1
=>(Pi/2)-1 ANSWER

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raviteja  (9 years ago)

can any one draw the figure?

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Anusiya .S.S  (9 years ago)

pi/2 -1
circle when divided into 4 equal part form sectors and one sector from each circle combined together form a square
combined area of 2 sectors =2(pi * r^2)/4
r=side of the square
so pi/2
area of square 1
so area common to both the circle is the curved form ie pi/2 -1

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