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

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 $A B C D)=72+56=128 \mathrm{~cm}^{2}$

 

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