Question

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

Answer 1

Area of $∆$ABD = $\sqrt{s(s-a)(s-b)(s-c)}$
$$\begin{gathered} s=\frac{1}{2}(a+b+c) \\ s=\frac{42+20+34}{2} \\ s=48\mathrm{cm} \end{gathered}$$

Area of $∆$ABD = $\sqrt{48(48-42)(48-20)(48-34)}$

$$\begin{gathered} =\sqrt{48\times 6\times 28\times 14} \\ =\sqrt{112896} \\ =336{\mathrm{cm}}^2 \end{gathered}$$

Area of $∆$BDC = $\sqrt{s(s-a)(s-b)(s-c)}$
$$\begin{gathered} s=\frac{1}{2}(a+b+c) \\ s=\frac{21+20+29}{2} \\ s=35\mathrm{cm} \end{gathered}$$

Area of $∆$BDC = $\sqrt{35(35-29)(35-20)(35-21)}$

$$\begin{gathered} =\sqrt{35\times 6\times 15\times 14} \\ =\sqrt{44100} \\ =210{\mathrm{cm}}^2 \end{gathered}$$

∴ Area of quadrilateral ABCD = Area of $∆$ABD + Area of $∆$BDC
​ = 336 + 210
= 546 cm²