Question

$OPQR$ is a square. $M,N$ are the midpoints of the sides $PQ$ and $QR$ respectively. If the ratio of the areas of the square and the $△OMN$ is $\lambda :6$, then $\frac{\lambda }{4}$ is equal to

• A. $2$
• B. $4$
• C. $8$
• D. $16$

Answer 1

The correct option is B $4$

Let us assume O be the origin
Therefore
$P=(a,0)$
$Q=(a,a)$
$R=(0,a)$
$M=(a,\frac{a}{2})$
$N=(\frac{a}{2},a)$

Therefore are of triangle $OMN$ is

$\frac{1}{2}\left|\begin{array}{cc}a& a/2\\ a/2& a\end{array}\right|$

$=\frac{1}{2}(a^2-\frac{a^2}{4})$

$=\frac{3a^2}{8}$

Taking ration of the areas of square to that of the triangle, we get

$\frac{a^2}{\frac{3a^2}{8}}$

$\frac{8}{3}=\frac{\lambda }{6}$

$\Rightarrow \lambda =16$

Hence $\frac{\lambda }{4}=4$