Latest Elitmus Aptitude Question SOLUTION: 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.

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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.

Asked In Elitmus vijay kumar mallik (8 years ago)

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abhinaw kumar  (8 years ago)

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..

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shashank garg  (8 years ago)

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

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Tushar Gupta  (8 years ago)

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

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Ankit  (8 years ago)

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.

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dytort  (8 years ago)

Why don't anyone post right answers on this website?

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Rahul Navariya  (8 years ago)

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

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Neeraj Shukla  (8 years ago)

here r=1 / root 2

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shailendra singh  (8 years ago)

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

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mohit  (8 years ago)

Anyone remember the options?

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abhilasha  (8 years ago)

Ans -96.56

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ananga  (8 years ago)

@abhinav...how got..
40*{1/(1-(1/(square root of 2)))} = 80+40(square root of 2)

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PRATIMA   (8 years ago)

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

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Adil khan  (6 years ago)

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

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