Starting with a unit square, a sequence of square is generated. Each square in the sequence has half the side length of its predecessor and two of its sides bisected by its predecessor's sides as shown. This process is repeated indefinitely. The total area enclosed by all the squares in limiting situation, is

Question

Starting with a unit square, a sequence of square is generated. Each square in the sequence has half the side length of its predecessor and two of its sides bisected by its predecessor's sides as shown. This process is repeated indefinitely. The total area enclosed by all the squares in limiting situation, is (A)  (5/4) sq. units (B) (79/64) sq. units (C) (75/64) sq. units (D) (1/12) sq. units

Tardigrade Admin · Accepted Answer

Required area =1+(3/4)[(1/4)+(1/42)+(1/43)+ ldots ldots ∞]=1+(3/4) ⋅ ((1/4)/1-(1)4)=1+(1/4)=(5/4) ⇒( A )

$\frac{5}{4}$ sq. units

$\frac{79}{64}$ sq. units

$\frac{75}{64}$ sq. units

$\frac{1}{12}$ sq. units

Required area $= 1 + \frac{3}{4} [\frac{1}{4} + \frac{1}{4 ^{2}} + \frac{1}{4 ^{3}} + \dots\dots \infty] = 1 + \frac{3}{4} \cdot \frac{\frac{1}{4}}{1 - \frac{1}{4}} = 1 + \frac{1}{4} = \frac{5}{4} \Rightarrow (A)$

45​ sq. units

6479​ sq. units

6475​ sq. units

121​ sq. units

Required area =1+43​[41​+421​+431​+……∞]=1+43​⋅1−41​41​​=1+41​=45​⇒(A)

$\frac{5}{4}$ sq. units

$\frac{79}{64}$ sq. units

$\frac{75}{64}$ sq. units

$\frac{1}{12}$ sq. units

Required area $= 1 + \frac{3}{4} [\frac{1}{4} + \frac{1}{4 ^{2}} + \frac{1}{4 ^{3}} + \dots\dots \infty] = 1 + \frac{3}{4} \cdot \frac{\frac{1}{4}}{1 - \frac{1}{4}} = 1 + \frac{1}{4} = \frac{5}{4} \Rightarrow (A)$