Side of a square is 10 cm. after joining the mid point of all sides makes a another inner square and this process goes to infinite.Find the sum of perimetere of all squares.

• perimeter of largest square= 4 *(10)=40
  perimeter of 2nd largest square = 4*(square root of 2)*5=20 square root of 2=40/(square root of 2)
  ------ 3rd -------------------=4*5=20= 40/(square root of 2)^2
  so on.................
  so , sum of all perimeter is = 40+40/(square root of 2) + 40/(square root of 2)^2 + .....
  = 40{1+1/(square root of 2)+1/(square root of 2)^2+......}
  this is G.P series with ratio is 1/(square root of 2)
  so, sum = 40*{1/(1-(1/(square root of 2)))} = 80+40(square root of 2) ans..
• 8 years agoHelpfull: Yes(37) No(0)
• This will form GP series 40, 20√2, 20, 10√2, 10 ....... and sum of n terms of infinite series is s=a/(1-r).
  Here r=1/√2 and a=40. putting the values we get sum = 136.568 cm
• 8 years agoHelpfull: Yes(12) No(2)
• sum=4(a+a/2+a/4+a/8......){where a=10}
  Its a Geometric Progression
  sum=a/(1-r) {here r=1/2 }
  sum=4*(10/(1/2))
  sum=4*20
  sum=80
  
• 8 years agoHelpfull: Yes(4) No(10)
• Answer will be=80+40(2)^1/2
  Explanation:Side of largest square=10cm therefore perimetre=4*side=40
  Side of second square which is formed by midpoints of side of square given =5(2)^1/2 by using pythogoras theorem, perimeter of this square will be=20(2)^1/2.
  In the same way series of Perimeter of squares will be= 40,20(2)^1/2, 20
  Which is a Geometric progression series with (2)^-1/2 as ratio difference
  Sum of gp for those series is =(a)/(1-r)= (40)/(1-1/(2)^1/2)= 80+ 40(2)^1/2.
• 8 years agoHelpfull: Yes(3) No(0)
• Why don't anyone post right answers on this website?
• 8 years agoHelpfull: Yes(2) No(5)
• side a=10cm
  joining the mid point of all sides makes a another inner square and this process goes to infinite
  sum of perimetere of all squares is = 4a+4a/2+1/2(4a/2)+......................
  so sum=4(a+a/2+a/4+a/8......)
  Its a infinite Geometric series
  sum=a/(1-r) {here r=1/2 }
  sum=4*(10/(1-1/2))
  sum=4*20
  sum=80
  
  
  
  
• 8 years agoHelpfull: Yes(1) No(12)
• here r=1 / root 2
  
• 8 years agoHelpfull: Yes(1) No(0)
• the correct solution is (40*root(2))/(root(2)-1)
  because there will be an infinite series of......40, (50)^1/2*4, (50/2)^1/2*4, (50/4)^1/2*4 and so on..
  sum of infinite gp= a/1-r
• 8 years agoHelpfull: Yes(0) No(0)
• Anyone remember the options?
  
• 8 years agoHelpfull: Yes(0) No(0)
• Ans -96.56
• 8 years agoHelpfull: Yes(0) No(0)
• @abhinav...how got..
  40*{1/(1-(1/(square root of 2)))} = 80+40(square root of 2)
  
• 8 years agoHelpfull: Yes(0) No(0)
• the perimeter of the 1st square will be 40 cm then next one would be half of it means 20cm n so on. .therefore sum of this infinite serious will be given by 40/(1-1/2) = 80 cm
• 8 years agoHelpfull: Yes(0) No(2)
• series will be
  40+20root2+20+10root2+.....
  easy way to seprate them into two series
  [40+20+10....]+[20root2+10root...] as u see r is 1/2 in both sries
  so 40/1-1/2 + 20root2/1-1/2
  80+40root2
• 6 years agoHelpfull: Yes(0) No(0)