Geometry & Mensuration
Aptitude

 Back to Questions
Q.

In the figure given below \(MN\) and \(NL\) are adjacent sides of a square and the arc \(MPL\) is drawn with \(N\) as centre and \(MN\) as radius. \(P\) is a point on the arc and \(PQRS\) is a square such that, \(RS\), if extended, passes through \(L\) while \(RQ\), if extended, passes through \(M\). What is the ratio of the area of a square of side \(MN\) and the square \(PQRS\)?

Image for Mensuration, Aptitude:2171-1

 A.

\(1:(3-2\sqrt{2})\)

 B.

\(2:(3-2\sqrt{2})\)

 C.

\(3:(3-2\sqrt{2})\)

 D.

\(4:(3-2\sqrt{2})\)

 Hide Ans

Solution:
Option(B) is correct

Answer image for Mensuration, Aptitude:2171-1

It  is  given  that  \(RS\)  passes  through  \(L\)  and  \(RQ\) passes through \(M\).  
 On  drawing  the  line  \(RSL\)  and  \(RQM\),  the diagram  shown  above  results.  From  the diagram,  it  is  clear  that \( MNLR\)  has  to  be  a square and \(P\) has to be the mid point of the arc.  
                  
 As  \(P \) is  the  mid-point  of  the  arc,  \(RN\)  passes through \(P\)

\(RP=a(\sqrt{2}-1)\)

Tom explains the factor give above $(\sqrt{2}-1)$ in the comment.

where \(MN = PN = a\)

\(\Rightarrow PQ=\dfrac{RP}{\sqrt{2}}\)

\(=a(\sqrt{2}-1)/\sqrt{2}\)

\(MN^2:PQ^2=2:3-2\sqrt{2}\)

More explanation on the last step is provided by Anita in the comments section.


(4) Comment(s)


Tushar
 ()

Can anyone explain the last step.


Anita
 ()

Last step,

$MN^2:PQ^2=2:3-2\sqrt{2}$

That is the ratio of the area for the larger square and the smaller square (as asked in the question).


Poonam Pipaliya
 ()

I don't get the logic behind $(\sqrt{2}−1)$


Tom
 ()

See, as $NP$ is radius, it length is $a$.

Now,

$RP=RN-NP$, where $RN$ is diagonal of the square with length $=\sqrt{a^2+a^2}=a\sqrt{2}$ and $NP=a$

Putting values,

$RP=RN-NP$

$RP=a\sqrt{2}-a$

$RP=a(\sqrt{2}-1)$

if (defined ( 'LF_SITECTRL' )) echo LF_SITECTRL; ?>