Question

Let $OPQR$ be a square and $M$ and $N$ be the midpoints of the sides $PQ$ and $QR$ respectively. The ratio of the area of square to the triangle $OMN$ is

• A. $4:1$
• B. $2:1$
• C. $8:3$
• D. $7:3$

Answer 1

The correct option is C $8:3$

Let the coordinates of vertices be
$O=(0,0)P=(a,0)Q=(a,a)R=(0,a)$
Now, the coordinates of $M$ and $N$ are
$M=(\frac{a+a}{2},\frac{0+a}{2})=(a,\frac{a}{2})N=(\frac{a}{2},a)$

Therefore, the area of $△OMN$
$=\frac{1}{2}\left|\begin{array}{cccc}{x}_1& x_2& x_3& x_1\\ y_1& y_2& y_3& y_1\end{array}\right|=\frac{1}{2}\left|\begin{array}{cccc}0& a& \frac{a}{2}& 0\\ 0& \frac{a}{2}& a& 0\end{array}\right|=\frac{1}{2}|0+(a^2-\frac{a^2}{4})+0|=\frac{3a^2}{8}\text{sq. units}$

Area of the square is $a^2$ sq. units.
Hence, the required ratio is $8:3$