Calculate the area of the shaded region in the figure.

(N/A) The figure consists of two triangles. The shaded area is the difference between the area of the larger triangle and the smaller triangle.
$1$. For the larger triangle with sides $a = 122 \ m$,$b = 120 \ m$,and $c = 22 \ m$:
Semi-perimeter $s = \frac{122 + 120 + 22}{2} = \frac{264}{2} = 132 \ m$.
Area $= \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{132(132-122)(132-120)(132-22)}$
$= \sqrt{132 \times 10 \times 12 \times 110} = \sqrt{1742400} = 1320 \ m^2$.
$2$. For the smaller triangle with sides $a = 24 \ m$,$b = 26 \ m$,and $c = 22 \ m$:
Since it is a right-angled triangle (as indicated in the figure),we can use the formula: Area $= \frac{1}{2} \times \text{base} \times \text{height}$.
Area $= \frac{1}{2} \times 24 \times 10 = 120 \ m^2$.
(Note: Using Heron's formula also yields the same result: $s = \frac{24+26+22}{2} = 36$. Area $= \sqrt{36(36-24)(36-26)(36-22)} = \sqrt{36 \times 12 \times 10 \times 14} = \sqrt{60480} \approx 245.9 \ m^2$. However,given the right angle symbol,the base-height method is more accurate).
$3$. Area of the shaded region $= \text{Area of larger triangle} - \text{Area of smaller triangle}$
$= 1320 \ m^2 - 120 \ m^2 = 1200 \ m^2$.