Question -
Answer -
Let ABCD be the given field in the form of a trapezium in which AB = 25 m and CD = 10 m, BC = 13 m, AD = 14 m and DC || AB.
Through C, draw CE||DA and let it meets AB at E.
Let CP h m be the height of the trapezium.
Now, DC || AE [∵ DC 11 AB given]
and CE||DA [by construction]
Thus, AECD is a parallelogram.
⇒ AE = DC = 10 m and CE = DA = 14 m
In A CEB, we have, CB = 13m, CE = 14 m
and BE = AB – AE = 25 -10 = 15 m
Let a = 14 m, b = 13 m and c = 15 m