For $△BCD$,
$s=\frac{25+17+12}{2}=27$
Using Heron's formula,
$A=\sqrt{27(27-25)(27-17)(27-12)}$
$=\sqrt{27\times 2\times 10\times 15}$
$=\sqrt{3\times 3\times 3\times 2\times 2\times 5\times 3\times 5}$
$=3\times 3\times 2\times 5$
$=90c{m}^2$
The parallelogram can also be divided into 2 triangles by diagonal AC, as shown below, the area of each triangle being $=90c{m}^2$.
Area of parallelogram $=90\times 2=180c{m}^2$

In, $△ACD$
Area of $△\text{ACD}=\frac{1}{2}\times \text{base}\times \text{Altitude}$
$90=\frac{1}{2}\times 12\times \text{Altitude}$
$\Rightarrow \text{Altitude}=15cm$