to the risk of sounding redundant, as the previous posting is absolutely correct.
Check the picture below.
we know the circum-inscribing figure is a square, namely a 23x20, so it has an area of 23*20 = 460 ft².
the distance from the bottom up, is 20, however to the tip of the triangle is 12, so the diffference is 20 - 12 = 8, meaning the triangle, in yellow, has a height/altitude of 8 ft, and we know its base is 23 ft.
so if we simply take the whole area of the square, and then subtract the area of the triangle, we're in effect making a whole in the squarish area, and what's leftover is the shaded area.
[tex]\bf \stackrel{\textit{square's area}}{(23\cdot 20)}~~~~-~~~~\stackrel{\textit{triangle's area}}{\cfrac{1}{2}(23)(8)}\implies 460~~-~~92\implies 368~ft^2[/tex]
