Let $ABCD$ be a trapezium with $AD$ parallel to $BC$. Assume there is a point $M$ in the interior of the segment $BC$ such that $AB=AM$ and $DC=DM$. Then,the ratio of the area of the trapezium to the area of $\triangle AMD$ is

The sides of a triangle are $3x+2y-6=0$,$2x-3y+6=0$,and $x+2y+2=0$. If $P(0, b)$ lies either on the triangle or inside the triangle,then $b$ lies in the interval

$A$ triangle is formed by the $Y$-axis,the straight line $L$ passing through the points $(3,0)$ and $(1, 4/3)$,and the straight line perpendicular to the line $L$ and passing through the point $(8,1)$. The area of the triangle (in sq units) is

If the points of intersection of the coordinate axes and $|x+y|=2$ form a rhombus,then its area is

The equation of a straight line,perpendicular to $3x - 4y = 6$ and forming a triangle of area $6 \text{ sq. units}$ with the coordinate axes,is

Let $A, B$ be points on the two half-lines $x - \sqrt{3}|y| = \alpha, \alpha > 0$ at a distance of $\alpha$ from their point of intersection $P$. The line segment $AB$ meets the angle bisector of the given half-lines at the point $Q$. If $PQ = \frac{9}{2}$ and $R$ is the radius of the circumcircle of $\triangle PAB$, then $\frac{\alpha^2}{R}$ is equal to . . . . . .