A quadrilateral field ABCD has AB = 9 $m$, BC = 40 $m$, CD = 28 $m$ and DA = 15 $m$, with a right angle at B. Find its area in $m^2$.

Join AC. In right triangle ABC, AC = √$(9^2+40^2)=$ √1681 = 41 m and its area = ½ × $9\times 40=180{\text{m}}^2$. In triangle ACD the sides are 41, 28 and 15 m, so $s=42$ and the area = √$(42\times 1\times 14\times 27)$ = √$15876=126{\text{m}}^2$. Total $=180+126=306{\text{m}}^2$.