Question

Find the area of quad. ABCD in which AB = 42 cm, BC = 21 cm, CD = 29 cm, DA = 34 cm and diag. BD = 20 cm.

Answer 1

Quadrilateral ABCD is divided into triangles $∆$ABD and $∆$BCD.
We will now use Hero's formula.
For $∆$ABD:
$\mathrm{Semiperimeter},s=\frac{1}{2}\left(42+20+34\right)=\frac{96}{2}=48\mathrm{cm}$

$$\begin{gathered} \mathrm{Area}\mathrm{of}∆\mathrm{ABD}=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{48\left(48-42\right)\left(48-34\right)\left(48-20\right)} \\ =\sqrt{48\times 6\times 14\times 28} \\ =\sqrt{112896} \\ =336{\mathrm{cm}}^2 \end{gathered}$$

For $∆$BCD:
$s=\frac{1}{2}\left(20+21+29\right)=\frac{70}{2}=35\mathrm{cm}$

$$\begin{gathered} \mathrm{Area}of∆\mathrm{BCD}=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{35\left(35-20\right)\left(35-21\right)\left(35-29\right)} \\ =\sqrt{35\times 15\times 14\times 6} \\ =\sqrt{44100} \\ =210{\mathrm{cm}}^2 \end{gathered}$$

Thus, we have:
Area of quadrilateral ABCD$=\mathrm{Ar}\left(∆ABD\right)+\mathrm{Ar}\left(BDC\right)=336+210=546{\mathrm{cm}}^2$