A triangle and a parallelogram have the same base and the same area.

The sides of the triangle DCE are

DC = 15 cm,

CE = 13 cm,

ED = 14 cm

Let the h be the height of parallelogram ABCD

Now, for the area of triangle DCE

Perimeter = DC + CE + ED

2s = 15 cm + 13 cm + 14 cm

s = 21 cm

By using Heron's Formula,

Area of the triangle $\mathrm{AOB}=\sqrt{\mathrm{s} \times(\mathrm{s}-\mathrm{a}) \times(\mathrm{s}-\mathrm{b}) \times(\mathrm{s}-\mathrm{c})}$

$=\sqrt{21 \times(7) \times(8) \times(6)}$

$=84 \mathrm{~cm}^{2}$

Also, area of triangle DCE $=$ Area of parallelogram $\mathrm{ABCD} \Rightarrow 84 \mathrm{~cm}^{2}$

$24 \times h=84 \mathrm{~cm}^{2}$

h = 6 cm.