find the area of quadrilateral ABCD in which AB=16cm, BC=CD=26cm, AD=12cm and the diagonal BD=20cm

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$$\begin{gathered} \mathrm{In} △\mathrm{ABD}, \\ \mathrm{Using} \mathrm{Heron}\text{'}\mathrm{s} \mathrm{Formula} \\ \mathrm{a}=16 \mathrm{cm} , \mathrm{b}=20 \mathrm{cm} , \mathrm{c}=12 \mathrm{cm} \\ \mathrm{s}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}=\frac{16+20+12}{2} \\ =\frac{48}{2}=24 \mathrm{cm} \\ \mathrm{Area} \left( △\mathrm{ABD} \right)=\sqrt{ \mathrm{s} \left(\mathrm{s}-\mathrm{a}\right) \left(\mathrm{s}-\mathrm{b}\right) \left(\mathrm{s}-\mathrm{c}\right)} \\ =\sqrt{24 \left(24-16\right) \left(24-20\right) \left(24-12\right)} \\ =\sqrt{ 24 \times 8 \times 4 \times 12} \\ =\sqrt{3\times 8\times 8\times 4\times 3\times 4} \\ =3\times 4\times 8 \\ =96 {\mathrm{cm}}^2. \\ \mathrm{In} △\mathrm{BCD}, \\ \mathrm{Using} \mathrm{Heron}\text{'}\mathrm{s} \mathrm{Formula} \\ \mathrm{a}=26 \mathrm{cm} , \mathrm{b}=20 \mathrm{cm} , \mathrm{c}=26 \mathrm{cm} \\ \mathrm{s}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}=\frac{26+20+26}{2} \\ =\frac{72}{2}=36 \mathrm{cm} \\ \mathrm{Area} \left( △\mathrm{BCD} \right)=\sqrt{ \mathrm{s} \left(\mathrm{s}-\mathrm{a}\right) \left(\mathrm{s}-\mathrm{b}\right) \left(\mathrm{s}-\mathrm{c}\right)} \\ =\sqrt{36 \left(36-26\right) \left(36-20\right) \left(36-26\right)} \\ =\sqrt{ 36 \times 10 \times 16 \times 10} \\ =6\times 10\times 4 \\ =240 {\mathrm{cm}}^2. \\ \mathrm{Area} \mathrm{of} \mathrm{quadrilateral} \mathrm{ABCD}=\mathrm{Area} \left( △\mathrm{ABD} \right) + \mathrm{Area} \left( △\mathrm{BCD} \right) \\ =96+240 \\ =336 {\mathrm{cm}}^2 \mathrm{ANS}... \end{gathered}$$

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