Page 6 - ln1

Let the given field is in the form of a trapezium ABCD such that parallel sides are AB =
10 m and DC = 25 m
Non-parallel sides are AD = 13 m and BC = 14 m.
We draw BE || AD, such that BE = 13 m.

The given field is divided into two shapes (i) ∆BCE, (ii) parallelogram ABED

For ∆BCE:
Sides of the triangle are a = 13 m, b = 14 m, c = 15 m

Let the height of the ∆BCE corresponding to the side EC be ‘h’ m.
Area of a triangle = x base x height

∴ x 15 x h = 84
⇒ h = 84×2 = m
15
1
Now, area of a trapezium = × + × h
2
1 56 1 56
2
= × 10 + 25 × = × 35 × = 196
2 5 2 5
So, area of the field
2
= 196

6