Two medians drawn from acute angles of a right angled triangles intersect at an angle of `pi//6`. If the length of the hypotenuse of the triangle is 3 units, then find the area of the triangle.

Explanation

In `Delta ABD, AD^(2) = c^(2) + a^(2)//4`
In `Delta FBC, CF^(2) = c^(2)//4 + a^(2)`

In `Delta AGF`, using cosine rule, we get
`cos 30^(@) = (AG^(2) + FG^(2) -(c^(2))/(4))/(2AG.FG)`
`:. (sqrt3)/(2) = (((2)/(3) AD)^(2) + ((1)/(3) CF)^(2) - (c^(2))/(4))/(2AG.FG)`
`rArr sqrt3 AG. FG = (4)/(9) (c^(2) + (a^(2))/(4)) + (1)/(9) (a^(2) + (c^(2))/(4)) - (c^(2))/(4)`
`rArr sqrt3 AG.FG = (2)/(9) (a^(2) + c^(2)) = 2 " " ("As " a^(2) + c^(2) = 9)`
`rArr sqrt3 AG.FG = 2`
`rArr AG.FG = (2)/(sqrt3)`
`:.` Area of `Delta ABC = 6 xx ("Area of " Delta AGF)`
`= 6 xx (1)/(2) AG.FG . sin 30^(@)`
`= 6 xx (1)/(2) xx (2)/(sqrt3) xx (1)/(2) = sqrt3` sq. units