Question:
In a quadrilateral ABCD, it is given that BD = 16 cm. If AL ⊥ BD and CM ⊥ BD such that AL = 9 cm and CM = 7 cm, then ar(quad. ABCD) = ?
(a) $256 \mathrm{~cm}^{2}$
(b) $128 \mathrm{~cm}^{2}$
(c) $64 \mathrm{~cm}^{2}$
(d) $96 \mathrm{~cm}^{2}$
Solution:
(b) $128 \mathrm{~cm}^{2}$
ar(quad ABCD) = ar (∆ ABD) + ar (∆ DBC)
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 $A B C D)=72+56=128 \mathrm{~cm}^{2}$