In figure, ΔABC has sides AB = 7.5 cm,

Solution:

Now, first determine the area of $\triangle A B C$.

The sides of a triangle are

$A B=\mathrm{a}=7.5 \mathrm{~cm}, \mathrm{BC}=\mathrm{b}=7 \mathrm{~cm}$ and $C A=c=6.5 \mathrm{~cm}$

Now, semi-perimeter of a triangle,

$s=\frac{a+b+c}{2}=\frac{7.5+7+6.5}{2}=\frac{21}{2}=10.5 \mathrm{~cm}$

$\therefore$ Area of $\triangle A B C=\sqrt{s(s-a)(s-b)(s-c)}$ [by Heron's formula]

$=\sqrt{10.5(10.5-7.5)(10.5-7)(10.5-6.5)}$

$=\sqrt{10.5 \times 3 \times 3.5 \times 4}=\sqrt{441}=21 \mathrm{~cm}^{2}$  $\ldots$ (i)

Now, area of parallelogram $B C E D=$ Base $\times$ Height

$=B C \times D F=7 \times D F$ .....(ii)

According to the question,

Area of $\triangle A B C=$ Area of parallelogram $B C E D$

$\Rightarrow$ $21=7 \times D F$ [from Eqs. (i) and (ii)]

$\Rightarrow$ $D F=\frac{21}{4}=3 \mathrm{~cm}$

Hence, the height of parallelogram is $3 \mathrm{~cm}$.