The sides of a triangle are $35 \text{ cm}$,$54 \text{ cm}$,and $61 \text{ cm}$,respectively. The length of its longest altitude is:

Find the area of a triangle,two sides of which are $15\,cm$ and $28\,cm$ and the perimeter is $84\,cm$. (in $,cm^2$)

The perimeter of an isosceles triangle is $32 \, cm$. The ratio of the equal side to its base is $3: 2$. Find the area of the triangle.

$PQRS$ is a square. If $PR = 10 \, cm$,then the area of $PQRS = \dots \, cm^2$.

In quadrilateral $ABCD$, one of its diagonals $AC$ measures $20 \, cm$. The altitudes on $AC$ from vertices $B$ and $D$ are $8 \, cm$ and $12 \, cm$ respectively. Find the area of quadrilateral $ABCD$ in $cm^2$.

In the figure,$\Delta ABC$ has sides $AB = 7.5\, cm$,$AC = 6.5\, cm$,and $BC = 7\, cm$. On base $BC$,a parallelogram $DBCE$ of the same area as that of $\Delta ABC$ is constructed. Find the height $DF$ of the parallelogram (in $cm$).