In a `Delta A B C ,\ A B=15 c m ,\ B C=13 c m\ a n d\ A C=14 c mdot` Find the area of ` Delta A B C\ a n d\ ` hence its altitude on `A C`

Explanation

Solution
Area of triangle= `sqrt(s(s−a)(s−b)(s−c))`
where s is the semi-perimeter
Here, `s= 2/(13+14+15)=21`
Therefore, area= `sqrt(21(6)(7)(8))= sqrt7056=84units^2`
Using Area= `2/1'×base×height`:
`84= 2/1×14×h⟹h=12 cm`