1Q.1 for circle of radius r , the side of square in it is r√2
from this
area of first circle is ∩r2-2r2
area of second circle is 2∩r2-4r2
difference is ∩r2-2r2
since difference between areas is same
sum= ∩r2-2r2+(n-1)∩r2-2r2
=n(∩r2-2r2)
Inscribed in a circle of radius R is square,a circle is inscribed in the square,a new square in the circle and so on for n times....
Q1 Sum of areas of all circles
Q2Limit of sum of areas of all squares as n→∞
Q3The limit of sum of areas of all circle as n→∞
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4 Answers
1
106Q1. R1 = R
R2 = R/√2
R3 = R/2
R4 = R/2√2
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.Rn = R/(√2)n-1
So sum of all areas of circles = piR2[1+1/2+1/4+1/8+...1/2n-1]
= piR2*[1*(1-1/2n)/(1-1/2) = 2piR2*(1-1/2n]
Q3. limit of areas of circles as n--> ∞ = 2piR2
106S1 = R√2 (sides of all squares)
S2 = R
S3 = R/√2
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.
.
Sn = R*/(√2)n-2
So sum of areas of all squares = R2[2+1+1/2+....+1/2n-2]
= R2*[2*(1-1/2n)/(1-1/2)]
= 4R2*(1-1/2n)
Taking limit = 4R2
