Points P, Q, R and S are on sides AB, BC, CD and DA of square ABCD, re

Question

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

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yashikaaggarwal · Answer

Answer is C.
The side of square ABCD is 1+n
The area of square ABCD is (1+n)^2 ----(1)
The side of square PQRS is (1+n^2)
Using Pythagoras Theorem,
In triangle APS
(AS)^2+(AP)^2=(PS)^2
(1)^2+(n)^2=(PS)^2
(PS)^2=1+n^2
(PS)=√(1+n^2)
Now the area of square PQRS is {√(1+n^2)}^2 = (1+n^2) ----(2)

The ratio of area of PQRS to area of ABCD
(1+n^2) : (1+n)^2

Answer is C

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GMATinsight · Answer

Let, 
n = 3
and 
Side of Square = 4

https://www.youtube.com/watch?v=1bWRC2plIaI

AnirudhaS · Answer

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

satya2029 · Answer

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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Points P, Q, R and S are on sides AB, BC, CD and DA of square ABCD, re

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