Latest TCS Aptitude Question SOLUTION: Two identical circles intersect so thta 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 tw

• Placement Papers
• /
• TCS
• /
• Numerical Ability
• /
• Geometry

TCS Question Papers: TCS Interview Experience(90) TCS Company Profile

(#M40038769) TCS Numerical Ability question Keep an EYE

Two identical circles intersect so thta 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 two circles

Asked In TCS anuhya (9 years ago)

Unsolved Read Solution (5)

Is this Puzzle helpful?   (7)   (2) Submit Your Solution Geometry

Read Solution (5) :

karthikeyank  (9 years ago)

Ans:pi/2 -1
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

Is this solution Helpfull? Yes (18) | No (1)  

Niveditha Ravichandran  (9 years ago)

area of the square = 1 sq cm
area of the sector of a circle when angle is provided = pi r^2 * angle /360
for a square the angle at all vertices is 90 deg
so pi*(1)^2* 90 / 360
= pi/4 ( this is for one circle)
for 2 circles = pi/2

total area of the 2 sectors of the circle = pi/2 ( this includes overlapped area)
so overlapped area = area of 2 sectors of circle - area of square
= pi/2 - 1

Is this solution Helpfull? Yes (11) | No (2)  

ABHINAV DINGRE  (9 years ago)

π/2 -1
common area is reduced from area of quarter sector

Is this solution Helpfull? Yes (1) | No (1)  

shashi kumar  (9 years ago)

0.57 sq cm

Is this solution Helpfull? Yes | No (5)  

Raghav gupta  (9 years ago)

please tell me how to draw the figure i cant visualize the figure

Is this solution Helpfull? Yes | No   

Please Login to Read Solution.