In the given figure, ABCD is a parallelogram. Find the height of the parallelogram. 

As ΔABE and parallelogram ABCD lie on the same base AB and between the same parallels.

∴ ar (ABCD) = 2 × ar (ΔABE) … (1)

Let the height of the parallelogram be h.

For ΔABE, a = 20 cm, b = 24 cm, and c = 36 cm

Semi-perimeter of ΔABE, $\mathrm{s}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}=\frac{20+24+36}{2}=40 \mathrm{cm}$

∴ ar (ΔABE) =  $\sqrt{\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}$

$$\begin{gathered} =\sqrt{40(40-20)(40-24)(40-36)} \\ =160\sqrt{2} {\mathrm{cm}}^2 \end{gathered}$$

Using (1), we obtain

Area (ACBD) = $2\times 160\sqrt{2}=320\sqrt{2} {\mathrm{cm}}^2$ 

We know that area of the parallelogram is base × height.

$$\begin{gathered} 320\sqrt{2}=36\times \mathrm{h} \\ \Rightarrow \mathrm{h}=\frac{320\sqrt{2}}{36}=12.57 \mathrm{cm} \end{gathered}$$

Thus, the height of the parallelogram is 12.57 cm.