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Answer :
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=\(660 m^2\)