Find the area of a parallelogram given in the figure. Also, find the length of the altitude from vertex `A` on the side `DC`.

Explanation

Area of parallelogram `ABCD=2("Area of " triangle BCD)`
Now, the sides of a `triangleBCD` are `a =12 cm, b=17 cm` and `c=25 cm.`
`therefore"Semi-perimeter of "triangleBCD, s=(a+b+c)/(2)=(12+17+25)/(2)=(54)/(2)=27cm`
`therefore"Area of " triangleBCD=sqrt(s(s-a)(s-b)(s-c))" "["by Heron's formula"]`
`=sqrt(27(27-12)(27-17)(27-25))`
`=sqrt(27xx15xx10xx2)`
`=sqrt(9xx3xx3xx5xx5xx2xx2)`
`=3xx3xx5xx2=90cm^(2)`
From Eq.(i),
Area of parallelogram `ABCD=2xx90`
`=180 cm^(2)`
Let altitude of parallelogram be h.
Also, area of parallelogram `ABCD=BasexxAltitude`
`rArr" "180=Dcxxh`
`rArr" "180=12xxh`
`therefore" "h=(180)/(12)=15cm`
Hence, the area of parallelogram is 180 `cm^(2)` and the length of altitude is `15 cm.`