The area of trapezium field whose parallel sides are `25 cm, 13 cm` and other sides are `15 cm and 15 cm.`

Explanation

Let `ABCD` be the given trapezium whose `AB=25 cm, CD = 13, BC = 15 cm` and `AD = 15 cm.`
Draw `" "` `CE||AD`
Now, ADCE is a parallelogram in which AD|| CE and AE||CD.
`:." " AE=DC=13` cm,
and `BE=AB-AE=25-13=12 cm `
In `DeltaBCE, s=(15+15+12)/(2)=21 `cm
`:." "` Area of `DeltaBCE=sqrt(s(s-a)(s-b)(s-c))`
`=sqrt(21(21-15)(21-15)(21-12))`
`=sqrt(21xx6xx6xx9)=18 sqrt(21) "cm"^(2) " " ...(1)`
Let h be the height of `DeltaBCE`, then
Area of `DeltaBCE=(1)/(2)("base"xx"height")`
`=(1)/(2)xx12xxh=6h " " (.:' EB=12)" " ...(2)`
From (1) and (2) , we have
`6h=18sqrt(21)`
`h=3sqrt(21)`cm
The height trapezium ABCD is same as that of `DeltaBCE`.
`:. ` Area of trapezium `=(1)/(2) (AB+CD)xxh`
`=(1)/(2)(25+13)xx3sqrt(21)=57sqrt(21) "cm"^(2)`.