There is a square of side 2 unit. on each and every 4 corner a circle is drawn and in the middle of the square an another fifth circle is drawn which is touching all the other 4 circles . The radius of all 5 circles are same.

now how much the inner area is left in the SQUARE which is not covered by the all 5 circle...???


i) 4(1-pi) ii)4-sqrt(2)*pi
iii) 4-pi /sqrt(2) iv)None of the above


very very easy question in geometry section till now :-)

• diagona of square is=sqrt(2)* side
  so,2sqrt(2)=4r
  then r=1/sqrt(2)
  and there are 4 sector of circle at the corner of square and 1 whole circle in the middle of the square.
  so area of sector =pi*r*r*angle/360 angle=90
  =pi/8
  and area of 4 sector=(pi/8)*4=pi/2
  and area of circle is=pi/2
  so total area=pi
  and area of square=4
  so left inner area =4-pi
  so the ans is (d)
• 10 years agoHelpfull: Yes(8) No(0)
• none of above
  
• 10 years agoHelpfull: Yes(5) No(2)
• ASHUTOSH KUMAR Diagonal is not 4r
  it is 4r+2*[(2^1/2)-1]r
• 10 years agoHelpfull: Yes(2) No(2)
• 4-2pi.........none of the above
• 10 years agoHelpfull: Yes(1) No(0)
• crct ans is 4-2pi
  
• 10 years agoHelpfull: Yes(1) No(0)
• None of Above
  
• 10 years agoHelpfull: Yes(0) No(0)
• answer is a
• 10 years agoHelpfull: Yes(0) No(3)
• Area of square- (Area of 4 quarter circle + 1 circle area)
  All circle having same radius.
  Given side of square=2, radius of circle=1
  So left area which is not covered by 5 circle=2^2-(pi*1^2+pi*1^2)=4-2pi
  which is not match with any of the given option so answer will be
  iv)None of the above.
  
• 10 years agoHelpfull: Yes(0) No(0)