Question

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

• A. $2$
• B. $4$
• C. $12$
• D. $16$

Answer 1

The correct option is B $4$

Area of square $=a^2$

Area of $\Delta OMN=\frac{1}{2}\left\{0\right(\frac{a}{2}-a)+a(a-0)+\frac{a}{2}(0-\frac{a}{2}\left)\right\}$

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

$=\frac{1}{2}\times \frac{4a^2-a^2}{4}=\frac{3a^2}{8}$

$\frac{\text{Area of square}}{\text{Area of}\Delta OMN}=\frac{a^2}{\frac{3a^2}{8}}=\frac{8}{3}$

Given that the ratio of square $OPQR$ and $△OMN$ is $\frac{\lambda }{6}$,

Now

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

$\lambda =16$

Then,

$\frac{\lambda }{4}=\frac{16}{4}=4$