Latest TCS Aptitude Question SOLUTION: 20. Two identical circles intersect so that their centres, and the points at which they intersect, form a square of side 1 cm. The area in sq. cm of the portion that is common to t

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20. Two identical circles intersect so that their centres, and the points at which they intersect, form a square of side 1 cm. The area in sq. cm of the portion that is common to the twocircles is:
a. (π/2) – 1
b. 4
c. √2 – 1
d. √5

Asked In TCS lav (7 years ago)

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Read Solution (5) :

shreoshree Chakraborty  (7 years ago)

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.

Now,

Area of shaded region BD

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

Let's call the above equation as EQ1

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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indrakaran  (7 years ago)

c.root2-1

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Sowjanya a v  (7 years ago)

We have to find the area of the blue shaded one and double it to get the area common to the both. Now this can be calculated as Area of the sector OAB - Area of the Triangle OAB.
As OA and OB are perpendicular, area of the sector OAB = 90360π(1)290360π(1)2 = π4π4
Area of the triangle OAB = 12×1×112×1×1 = 1212
Area common to both = 2(π4−12)=π2−12(π4−12)=π2−1

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saurabh  (7 years ago)

a is correct answer

Is this solution Helpfull? Yes | No (1)  

Akanksha  (7 years ago)

area of common portion is 2{ (pi/4)-(1/2) }=pi/2-1

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