In figure, `triangle``ABC` has sides `AB` = `7.5` `cm`, `AC` = `6.5` `cm` and `BC`=`7` `cm`. On base `BC` a parallelogram `DBCE` of same area as that of `triangle` `ABC` is constructed. Find the height `DF` of the parallelogram.

Explanation

Now, first determine the area of `triangle` ABC.
The sides of a triangle are
AB=a=7.5 cm, BC=b= 7 cm and CA=c=6.5 cm
Now, semi-perimeter of a triangle,
`s=(a+b+c)/(2)=(7.5+7+6.5)/(2)=(21)/(2)=10.5 cm`
`therefore" ""Area of "triangleABC=sqrt(s(s-a)(s-b)(s-c))" "["by Heron's formula"]`
`=sqrt(10.5(10.5-7.5)(10.5-7)(10.5-6.5))`
`=sqrt(10.5xx3xx35xx4)=sqrt441=21cm^(2)`
Now, area of parallelogram `BCED="Base"xx"Height"`
`=BCxxDF=7xxDF`
According to the question,
Area of `triangleABC`= Area of parallelogram BCED
`rArr " "21=7xxDF" "["from Eqs.(i) and (ii)"]`
`rArr " "DF=(21)/(4)=3cm`
Hence, the height of parallelogram is 3 cm.