Two medians drawn from the acute angles of a right angled triangle intersect at an angle `pi/6dot` If the length of the hypotenuse of the triangle is `3u n i t s ,` then the area of the triangle (in sq. units) is `sqrt(3)` (b) 3 (c) `sqrt(2)` (d) 9

Explanation

In the figure, ABC is right angled triangle.
Given that `AC=3`
So, ` a^2+b^2=9`
AD and CF are two medians.
So, `D=((a)/(2),0)` and `F-=(0,(b)/(2))`
Slope of `AD=(-2b)/(a)`
Slope of `CF=(-b)/(2a)`
`therefore" " "tan"(pi)/(6)=|((-2b)/a+(b)/(2a))/(a+(b^2)/(a^2))|=|(-3ab)/(2(a^2+b^2))|=(3ab)/(9xx2)`
`rArr (1)/(sqrt3)=(ab)/(6)`
`rArrab=2sqrt(3)`
`therefore` Area of triangle `ABC =(1)/(2) ab=sqrt(3)`