Question:
The length of the fence of a trapezium-shaped field ABCD is 130 m and side AB is perpendicular to each of the parallel sides AD and BC. If BC = 54 m, CD = 19 m and AD = 42 m, find the area of the field..png)
Solution:
Length of the side $\mathrm{AB}=(130-(54+19+42)) \mathrm{m}$
$=15 \mathrm{~m}$
Area of the trapezium $-$ shaped field $=\left\{\frac{1}{2} \times(\mathrm{AD}+\mathrm{BC}) \times \mathrm{AB}\right\}$
$=\left\{\frac{1}{2} \times(42+54) \times 15\right\} \mathrm{m}^{2}$
$=\left(\frac{1}{2} \times 96 \times 15\right) \mathrm{m}^{2}$
$=(48 \times 15) \mathrm{m}^{2}$
$=720 \mathrm{~m}^{2}$
Hence, the area of the field is $720 \mathrm{~m}^{2}$.