A Geometry Problem

Note that the area of twice the gray region (a parallelogram) is obviously half the area of the rectangle. Therefore, the area of the gray region is half of half the area of the rectangle, or $120\div 4=\boxed{30\text{ m}^2}$ .

Let x and y be the length and the breadth of the rectangle respectively.So xy=120 _(1). Now 2 * (grey area)=a parallelogram's area _ _ (2) ,the area of the parallelogram=( x/2) * y(as M and N are midpoints)=xy/2 ,now putting it in (2), 2 * (area of grey area)=xy/2 or area of grey area=xy/4 or area of grey area=120/4 {from (1)} so area of grey area=30 m^2

If you cut the rectangle in half in the middle and move the little grey triangle sticking out on the left to the bottom part to complete the bigger triangle, you will notice that it results in a triangle that is 1/4 the area of the rectangle. Therefore, 120/4 = 30.

$I\quad proved\quad that\quad the\quad area\quad of\quad the\quad grey\quad region\quad is\quad \frac { 1 }{ 2 } \quad the\quad area\quad of\\ \quad the\quad paralellogram\quad in\quad the\quad middle\quad using\quad congruent\quad triangles\quad$