Latest TCS Aptitude Question SOLUTION: In an exam, students were asked to find the perimeter of a square which contains three circles such that their enters on the diagonal of the square , the middle circle touches two

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In an exam, students were asked to find the perimeter of a square which contains three circles such that their enters on the diagonal of the square , the middle circle touches two extreme circles on either side and two extreme circles touch two sides of the square and the radius of the three are equal . And ate what times the radius equals the perimeter?

Asked In TCS Nikhil (10 years ago)

Solved anu Read Solution (10)

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Read Solution (10) :

anu  (10 years ago)

Let the radius of circle be r
let the side of square be a
then diagonal of square= a*sqrt(2)
This diagonal length = 4*r + 2r * sqrt(2)
(because the extreme circle's radius is perpendicular to side of square.)
Thus we get

4*r+2r*sqrt(2)=a*sqrt(2)

r(2*sqrt(2)+2)=a

r/a=1/(2*sqrt(2)+2)

Thus ratio: r:4a = 1:4(2*sqrt(2)+2)

Is this solution Helpfull? Yes (20) | No (2)  

Saloni Singh  (10 years ago)

let side of sqr =a
perimeter=4*a
also diagonal=a* sqrt(2)
as the 3 circles lie on diagonal,so
6*r=a*sqrt(2)
substituting a= perimeter/4
6*r=p*sqrt(2)/4
12*sqrt(2) r=p

Is this solution Helpfull? Yes (13) | No (11)  

richa ahuja  (10 years ago)

side of square =a
no of circles=n=3(here)
radius of each circle=r
a:r=((n-1)sqrt(2)+2):1
solving this we get perimeter=(8*sqrt(2)+8)r

Is this solution Helpfull? Yes (4) | No (5)  

Amandeep Singh  (10 years ago)

32 times Radius = Parameter of Square.
COZ, ATQ,
Parameter = 2(3*Raduis(R) + Remaining part of Diagonal of Square(X)).
Solve for R=1 and X=1.
Side of Square (A) = 8 units.
hence,
32R = P.

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

Manish Kumar  (10 years ago)

12(root2) is ans

Is this solution Helpfull? Yes (2) | No (3)  

rahul ranjan  (10 years ago)

a(sqrt2) = diagonal = r sqrt2 + r + 2r + r + r sqrt2 = 4r + 2r sqrt2
a = (4r + 2r sqrt2) / sqrt2
a = 4r/sqrt2 + 2r sqrt2 / sqrt2
a = (2 x 2r) / sqrt2 + 2r

But we know, 2/sqrt2 = sqrt2. Therefore, above equation becomes,
Let the radius of circle be r.
Let the side of the square be a.
Then Diagonal = a(sqrt2) a = 2r sqrt2 + 2r = 2r(sqrt2+1)
Then perimeter of the square = 4a = 4 x 2r(sqrt2+1) = 8r(sqrt2+1)
= 8r(1.414+1) = 8r(2.414) = r(19.312)
Hence 19.312 time radius equals the perimeter of the square.

Is this solution Helpfull? Yes (2) | No   

juhe rawat  (10 years ago)

@nikhil
which is the right ans...?

Is this solution Helpfull? Yes (1) | No   

sourav  (10 years ago)

(sqrt(2)+1)*8..

side of square=(4r+sqrt(2)r+sqrt(2)r)/sqrt(2)

Is this solution Helpfull? Yes | No (1)  

rohin  (10 years ago)

see the sol of AMANDEEP SINGH its right

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Gaurav Kumar  (10 years ago)

2sqrt2(r)+4r=sqrt2 (a)
a=2r(1+sqrt2)
p=8r(1+sqrt2)

Is this solution Helpfull? Yes | No   

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