If in a triangle ABC, AB = 5 units, $$\angle B = {\cos ^{ - 1}}\left( {{3 \over 5}} \right)$$ and radius of circumcircle of $$\Delta$$ABC is 5 units, then the area (in sq. units) of $$\Delta$$ABC is :

The triangle of maximum area that can be inscribed in a given circle of radius 'r' is :

Let a, b, c be in arithmetic progression. Let the centroid of the triangle with vertices (a, c), (2, b) and (a, b) be $$\left( {{{10} \over 3},{7 \over 3}} \right)$$. If $$\alpha$$, $$\beta$$ are the roots of the equation $$a{x^2} + bx + 1 = 0$$, then the value of $${\alpha ^2} + {\beta ^2} - \alpha \beta $$ is :

A triangle ABC lying in the first quadrant has two vertices as A(1, 2) and B(3, 1). If $$\angle BAC = {90^o}$$ and area$$\left( {\Delta ABC} \right) = 5\sqrt 5 $$ s units, then the abscissa of the vertex C is :