The diagonals of a parallelogram $ABCD$ are along the lines $x+3y=4$ and $6x-2y=7$. Then $ABCD$ must be a

If a straight line perpendicular to $2x - 3y + 7 = 0$ forms a triangle with the coordinate axes whose area is $3 \text{ sq. units}$,then the equation of the straight line is:

Let $ABCD$ be a quadrilateral such that there exists a point $E$ inside the quadrilateral satisfying $AE=BE=CE=DE$. Suppose $\angle DAB, \angle ABC, \angle BCD$ are in an arithmetic progression. Then the median of the set $\{\angle DAB, \angle ABC, \angle BCD\}$ is

Suppose $ABCD$ $(AB \parallel CD)$ is a trapezium such that the diagonals $AC$ and $BD$ bisect the angles $\angle DAB$ and $\angle CBA$,respectively. Then

Let $ABC$ be an equilateral triangle,and let $KLMN$ be a rectangle with $K, L$ on $BC$,$M$ on $AC$,and $N$ on $AB$. Suppose $AN / NB = 2$ and the area of $\triangle BKN$ is $6$. The area of the $\triangle ABC$ is:

In a triangle $ABC$,$\angle BAC = 90^{\circ}$; $AD$ is the altitude from $A$ onto $BC$. Draw $DE$ perpendicular to $AC$ and $DF$ perpendicular to $AB$. Suppose $AB = 15$ and $BC = 25$. Then the length of $EF$ is