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Let the proportional constant be 1 (because we can!)
So \(area ABCD = (1+n)^2\)
And, \(Area PQRS = Area ABCD - 4*(∆ABS)\)
\(or, Area PQRS = (1+n)^2 - 4.\frac{1}{2}.1.n = (1+n)^2 - 2n\)

Therefore \(\frac{Area PQRS}{ Area ABCD} = \frac{(1+n)^2-2n}{(1+n)^2} = \frac{1+n^2}{(1+n)^2}\)

Answer: C
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Bunuel
Points P, Q, R and S are on sides AB, BC, CD and DA of square ABCD, respectively, so that AP : PB = BQ : QC = CR : RD = DS : SA = 1 : n. What is the ratio of the area of PQRS to the area of ABCD ?


A. \(1 : (1 + n)\)

B. \(1 : n\)

C. \((1 + n^2) : (1 + n)^2\)

D. \((1 + n ) : (1 + n^2)\)

E. \(n : 1\)


Are You Up For the Challenge: 700 Level Questions
Assume side of PQRS is 5 unit, this gives side of ABCD(3+4)=7 unit
Area required=25/49
here we get 1/n=3/4 or 1/n=4/3(Depend upon which side you choose)
From either of the values of n, C gives the area required 25/49
C:)
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