A right triangle is drawn in a semicircle of radius `1/2` with one of its legs along the diameter. If the maximum area of the triangle is `M` , then the value of `32sqrt(3)M c` is______

Answer

9

Explanation

`BCxxCe=ACxxCD`
`therfore (BC)(CE)=x(1-x)`
But BC=CE
`there BC =sqrt(x(1-x))`
`therefore Area triangle=xsqrt(x-x^(2))/(2)`
or `triangle=(x^(3)-x^(4))/(4)`
or `(d triangle^(2))/(dx)=(3x^(2)-4x^(3))/(4)`
If `(d triangle^(2))/(dx)=0` then `x=3/4` which is the point of maxima
Hence maximum area is `(3 sqrt(3))/(32)`