A field is shaped like a trapezium whose parallel sides are 31 m and 10 m, while the two remaining sides measure 13 m and 20 m. What is the area of the field, in square metres?

Drop a perpendicular of height $h$ from each end of the 10 m side onto the 31 m side. The two offcuts have bases summing to 31 - 10 = 21 m, so √$(13^2-h^2)$ + √$(20^2-h^2)=21$; $h=12$ works, since 5 + 16 = 21. The area = $(1/2)(31+10)\times 12=246$ sq m. The trap 492 drops the 1/2; $126=(1/2)\times 21\times 12$ uses the DIFFERENCE of the parallel sides where their sum belongs; $266.5=(1/2)\times 41\times 13$ mistakes the 13 m slanting side for the perpendicular height.