In a quadrilateral ABCD, it is given that BD = 16 cm.

(b) $128 \mathrm{~cm}^{2}$

ar(quad ABCD) = ar (∆ ABD) + ​ar (∆ DBC)
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We have the following:

$\operatorname{ar}(\Delta A B D)=\frac{1}{2} \times$ base $\times$ height $=\frac{1}{2} \times 16 \times 9=72 \mathrm{~cm}^{2}$

$\operatorname{ar}(\Delta D B C)=\frac{1}{2} \times$ base $\times$ height $=\frac{1}{2} \times 16 \times 7=56 \mathrm{~cm}^{2}$

$\therefore \operatorname{ar}($ quad $A B C D)=72+56=128 \mathrm{~cm}^{2}$