Find the area of the parallelogram given in the figure. Also,find the length of the altitude from vertex $A$ on the side $DC.$

(N/A) The sides of $\Delta BCD$ are $a = 12 \text{ cm}, b = 17 \text{ cm},$ and $c = 25 \text{ cm}.$
$\therefore$ The semi-perimeter $(s)$ of $\Delta BCD$ is:
$s = \frac{a + b + c}{2} = \frac{12 + 17 + 25}{2} = \frac{54}{2} = 27 \text{ cm}.$
$\therefore$ Area of $\Delta BCD = \sqrt{s(s - a)(s - b)(s - c)}$ [By Heron's formula]
$= \sqrt{27(27 - 12)(27 - 17)(27 - 25)}$
$= \sqrt{27 \times 15 \times 10 \times 2}$
$= \sqrt{(9 \times 3) \times (3 \times 5) \times (5 \times 2) \times 2}$
$= \sqrt{9 \times 9 \times 25 \times 4} = 3 \times 3 \times 5 \times 2 = 90 \text{ cm}^2.$
Since a diagonal divides a parallelogram into two triangles of equal area:
Area of parallelogram $ABCD = 2 \times \text{Area of } \Delta BCD = 2 \times 90 = 180 \text{ cm}^2.$
Let the altitude from vertex $A$ to side $DC$ be $h.$
Area of parallelogram $ABCD = \text{Base} \times \text{Altitude}$
$180 = DC \times h$
$180 = 12 \times h$
$h = \frac{180}{12} = 15 \text{ cm}.$
Hence,the area of the parallelogram is $180 \text{ cm}^2$ and the length of the altitude is $15 \text{ cm}.$