3.Length of the fence of a trapezium A shaped field ABCD is 120 m. IF BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.

Given:
BC = 48 m, CD = 17 m, AD = 40 m

AB + BC + CD + DA = 120 m .

AB = 120 m – (48 m + 17 m + 40 m) = 120 – 105 m = 15 m

So the area of trapezium fence ABCD=\(\frac{1}2\times (sum\; of \;parallel \;sides)\times (distances \;between\; them)\)

\(=\frac{1}2\times (BC+AD)\times AB\)

\(=\frac{1}2\times (48+40)\times 15\)

\(=\frac{1}2\times 88\times 15\)

\(=44\times 15=660m^2\)

So, the required area=\(660 m^2\)