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3 Length of the fence of a trapezium shaped field $ABCD$ is $120 m$ . If $BC = 48 m$ , $CD = 17 m$ and $AD = 40 m$ , find the area of this field. Side $AB$ is perpendicular to the parallel sides $AD$ and $BC$ .

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Given: $BC = 48 m$ , $CD = 17 m$ and $AD = 40 m$ ,

Length of the fence of a trapezium-shaped field,

$ABCD$ = $120 m$ = $AB+BC+CD+DA$

$120=AB+48+17+40$

$AB=15m$

Area of trapezium = $\frac{1}{2}\times sum \, of\, parallel\, sides\times height$

$=\frac{1}{2}\times (40+48)\times 15$

$=\frac{1}{2}\times (88)\times 15$

$= (44)\times 15$

$=660m^{2}$

Posted by

seema garhwal

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