The perimeter of a triangle is $16 \text{ cm}$,and one of the sides is of length $6 \text{ cm}$. If the area of the triangle is $12 \text{ cm}^2$,then the triangle is:

In the figure given below,$ABCDEF$ is a regular hexagon of side length $1$,$AFPS$ and $ABQR$ are squares. Then,the ratio $\frac{\operatorname{ar}(APQ)}{\operatorname{ar}(SRP)}$ equals

The incentre of the triangle formed by the lines $x = 0$,$y = 0$,and $3x + 4y = 12$ is

If $ad-bc \neq 0$,then the area (in sq. units) of the parallelogram formed by the lines $ax+by+2=0$,$ax+by+5=0$,$cx+dy+3=0$ and $cx+dy+7=0$ is

If $\Delta_1$ is the area of the triangle formed by the centroid and two vertices of a triangle,and $\Delta_2$ is the area of the triangle formed by the mid-points of the sides of the same triangle,then $\Delta_1 : \Delta_2 =$

The equations of two sides $AB$ and $AC$ of a triangle $ABC$ are $4x + y = 14$ and $3x - 2y = 5$,respectively. The point $\left(2, -\frac{4}{3}\right)$ divides the third side $BC$ internally in the ratio $2:1$. The equation of the side $BC$ is: