Let the lengths of the altitudes drawn from the vertices of `Delta ABC` to the opposite sides are 2, 2 and 3. If the area of `Delta ABC " is " Delta`, then find the area of triangle

Answer

`(9)/(2sqrt2)` sq. unit

Explanation

Given that `a = (2Delta)/(h_(A)) , b = (2Delta)/(h_(B)) , c = (2Delta)/(h_(C))`,
i.e., `a = Delta, b = Delta , c = (2Delta)/(3)`

`rArr a + b + c = 2 Delta ((1)/(h_(A)) + (1)/(h_(B)) + (1)/(h_(C)))`
or `2s = 2Delta ((1)/(2) + (1)/(2) + (1)/(3)) = 2Delta ((8)/(6))`
or `s = (4)/(3) Delta`...(i)
Now `Delta^(2) = s (s -a) (s-b) (s -c)`
or `Delta^(2) = (4)/(3) xx Delta ((Delta)/(3)) ((Delta)/(3)) (2Delta)/(3))`
or `1 = (8)/(81) xx Delta^(2)`
or `Delta = (9)/(sqrt8)`