In the given figure ABCD is a quadrilateral in which ∠ABC = 90°

Solution:

(b) 114 sq cm

Using Pythagoras' theorem in $\triangle A B C$, we get:

$A C^{2}=A B^{2}+B C^{2}$

$\Rightarrow A B=\sqrt{A C^{2}-B C^{2}}$

$=\sqrt{17^{2}-15^{2}}$

 

$=8 \mathrm{~cm}$

Area of $\Delta A B C=\frac{1}{2} \times A B \times B C$

$=\frac{1}{2} \times 8 \times 15$

$=60 \mathrm{~cm}^{2}$

Area of $\Delta B C D=\frac{1}{2} \times B D \times C D$

$=\frac{1}{2} \times 12 \times 9$

$=54 \mathrm{~cm}^{2}$

$\therefore$ Area of quadrilateral $A B C D=\operatorname{Ar}(\Delta A B C)+\operatorname{Ar}(\Delta B C D)=54+60=114 \mathrm{~cm}^{2}$