The area of quadrilateral ABCD in the given figure is


The area of quadrilateral ABCD in the given figure is
(a) 57 cm2
(b) 108 cm2
(c) 114 cm2
(d) 195 cm2



(c) $114 \mathrm{~cm}^{2}$

ar (quad. ABCD) =  ar (∆ ABC)  +  ar (∆ ACD)
In right angle triangle ACD, we have:

$A C=\sqrt{\left(17^{2}-8^{2}\right)}=\sqrt{225}=15 \mathrm{~cm}$

In right angle triangle ABC, we have:

$B C=\sqrt{\left(15^{2}-9^{2}\right)}=\sqrt{144}=12 \mathrm{~cm}$

Now, we have the following:

$\operatorname{ar}(\Delta A B C)=\frac{1}{2} \times 12 \times 9=54 \mathrm{~cm}^{2}$

$\operatorname{ar}(\Delta A D C)=\frac{1}{2} \times 15 \times 8=60 \mathrm{~cm}^{2}$

$\operatorname{ar}($ quad $. A B C D)=54+60=114 \mathrm{~cm}^{2}$


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