Find the area of the triangle with the length of the sides $9 \, cm$,$10 \, cm$,and $17 \, cm$. (in $, cm^2$)

Write True or False and justify your answer.
If $a, b, c$ are the lengths of three sides of a triangle, then area of a triangle $= \sqrt{s(s-a)(s-b)(s-c)}$, where $s = \text{perimeter of triangle}$.

If the sides of a triangular field are $290 \, m$,$290 \, m$,and $400 \, m$,find the area of the field in $m^2$.

The lengths of the sides of a triangle are $13 \, cm$,$13 \, cm$,and $10 \, cm$. Find its area. Also,find the length of the altitude from the opposite vertex to the side whose length is $10 \, cm$.

The area of an equilateral triangle with side $2 \sqrt{3} \, cm$ is (in $cm^2$):

In a parallelogram $ABCD$,the base $AB$ and the corresponding altitude $DM$ are $16 \, cm$ and $12 \, cm$,then the area of $ABCD =$ ................ $cm^{2}$.