The perimeter of a triangle is $16 \text{ cm}$, 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:

Let $a$ be the length of a side of a square $OABC$ with $O$ being the origin. Its side $OA$ makes an acute angle $\alpha$ with the positive $x$-axis and the equations of its diagonals are $(\sqrt{3}+1)x+(\sqrt{3}-1)y=0$ and $(\sqrt{3}-1)x-(\sqrt{3}+1)y+8\sqrt{3}=0$. Then $a^2$ is equal to

One of the vertices of a square is the origin,and the adjacent sides of the square lie along the positive $x$ and $y$ axes. If the side length is $5$,which of the following is $NOT$ a vertex of the square?

For what value of $k$ are the points $(k, 2 - 2k)$,$(1 - k, 2k)$,and $(-4 - k, 6 - 2k)$ collinear?

The area enclosed by the graphs of $|x + y| = 2$ and $|x| = 1$ is

If $x_1, x_2, x_3$ and $y_1, y_2, y_3$ are in $G.P.$ with the same common ratio,then the points $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ are: