Find the area of a trapezium whose parallel sides `25 c m ,13 c m` and other sides are `15 c ma n d15 c mdot`

Explanation

Let `ABCD` be the given trapezium in which,`AB=25cm,CD=13cm,BC=15cm andAD=15cm`
Draw`CE∥AD \and\ CF_|_BE`
Now, `ADCE` is a parallelogram in which `AD∥CE \and\ AE∥CD`
`AE=DC=13cm`
`BE=AB−AE=25−13=12cm`
In `/_\BCE, \we have\,s=1/2(15+15+12)=21cm`
Area of`/_\BCE=sqrt(s(s−a)(s−b)(s−c))`
`=> Area of /_\BCE=sqrt(21(21−15)(21−15)(21−12))`
`=sqrt(21×6×6×9`
`18sqrt21sq.cm`(i)

Let h be the height of `/_\BCE, then,Area of/_\BCE=1/2(BasexxHeight)`
`=1/2 ×12×h=6h`(ii)
From (i) and (ii), we have,`6h=18sqrt21`
`h=3
sqrt21cm`
Clearly, the height of trapezium `ABCD` is same as that of `/_\BCE`
Area of trapezium`=1/2 (AB+CD)×h`
Area of trapezium`=1/2(25+13)×3sqrt
21cm^2`
`=57sqrt21cm^2`