Consider $$\Delta$$ ABC
It is given that AB = 7.5cm, BC = 7cm and AC = 6.5cm
Take a = 7.5cm, b = 7cm and c = 6.5 cm
So we get
$$s=\dfrac{a+b+c}{2}$$
$$s=\dfrac{7.5+7+6.5}{2}$$
By division
s=10.5cm
We know that
Area=$$\sqrt{s(s-a)(s-b)(s-c)}$$
By substituting the values
Area=$$ \sqrt{10.5(10.5-7.5)(10.5-7)(10.5-6.5)}$$
So we get
Area=$$ \sqrt{10.5\times3\times3.5\times4}$$
It can be written as
Area=$$\sqrt{441}$$
By multiplication
Area=$$21cm^2$$
We know that the area of parallelogram DBCE = Area of $$\Delta$$ ABC
It can be written as
$$BC\times DL = 21$$
By substituting the values
$$7 \times DL = 21$$
By division
DL = 3cm
Therefore, the height DL of the parallelogra