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

Sides of triangle ABC are 7.5 cm, 7 cm and 6.5 cm.
the semi-perimeter of `DeltaABC`,
`s=(7.5+7+6.5)/(2)=(21)/(2)=10.5 cm`
Area of `Delta = sqrt(s(s-a)(s-b)(s-c))`
`=sqrt(10.5(10.5-7.5)(10.5-7)(10.5-6.5))`
`=sqrt(10.5xx3xx3.5xx4)=sqrt(31.5xx14)`
`=sqrt(441)=21 cm^(2)`
Now, as on base BC a parallelogram DBCE of same area as that of `DeltaABC` is constructed.
Therefore, area of parallelogram DBCE = 21 `cm^(2)`. Also, area of parallelogram `DeltaBCE=BCxxDF`.
`:. BCxxDF=12 cm^(2)`
`rArr 7xxDF=21 cm^(2)`
`rArr DF=21 cm^(2)div 7cm=3 cm`
Hence, the height DF of the parallelogram = 3 cm.