The length of the fence of a trapezium-shaped field $ A B C D $ is $ 130 \mathrm{~m} $ and side $ A B $ is perpendicular to each of the parallel sides AD and $ B C $. If $ B C=54 \mathrm{~m}, C D=19 \mathrm{~m} $ and $ A D=42 \mathrm{~m} $, find the area of the field.

Given:

The length of the fence of a trapezium-shaped field \( A B C D \) is \( 130 \mathrm{~m} \) and side \( A B \) is perpendicular to each of the parallel sides AD and \( B C \). 

\( B C=54 \mathrm{~m}, C D=19 \mathrm{~m} \) and \( A D=42 \mathrm{~m} \).

To do:

We have to find the area of the field.

Solution:

From the figure,

$BC=54\ m, CD=19\ m, AD=42\ m$

Perimeter$=$Length of the fence

$130\ m=AB+BC+CD+DA$

$130=AB+54+19+42$

$AB=130-115$

$AB=15\ m$

Area of the field$=\frac{1}{2}\times$Sum of parallel sides $\times$Perpendicular distance(AB) 

$=\frac{1}{2}\times(42+54)\times15\ cm^2$

$=\frac{1}{2}\times96\times15\ cm^2$

$=48\times15\ cm^2$

$=720\ cm^2$