# Difficult Geometry & Mensuration Solved QuestionAptitude Discussion

 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 $PQRS$?
 ✖ A. $1:(3-2\sqrt{2})$ ✔ B. $2:(3-2\sqrt{2})$ ✖ C. $3:(3-2\sqrt{2})$ ✖ D. $4:(3-2\sqrt{2})$

Solution:
Option(B) is correct

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 $P$

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

where $MN = PN = a$

$\Rightarrow PQ=\dfrac{RP}{\sqrt{2}}$

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

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

## (2) Comment(s)

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,

$RP=RN-NP$

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

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