What is the length of the altitude from vertex C on the side AB of a parallelogram ABCD?

 

In $∆\mathrm{BDC}$, DC = 24 cm, BC = 34 cm, BD = 50 cm

Semi-perimeter, $s=\frac{24+34+50}{2}=54$

From Heron's formula,
Area of triangle = $\sqrt{\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}$

Area of $∆\mathrm{BDC}=\sqrt{54(54-24)(54-34)(54-50)}$

$=360 c{m}^2$

Area of parallelogram ABCD = 2$\times$Area of $∆\mathrm{BDC}$

$$\begin{gathered} =2\times 360 \\ =720 {\mathrm{cm}}^2 \end{gathered}$$
Let the altitude of the parallelogram be h.
Also, area of parallelogram ABCD = Base × Altitude 
⇒ 720 = 24 × h
⇒ h = 30 cm