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


In the given figure ABCD is a quadrilateral in which ∠ABC = 90°, ∠BDC = 90°, AC = 17 cm, BC = 15 cm, BD = 12 cm and CD = 9 cm. The area of quad. ABCD is

(a) 102 cm2
(b) 114 cm2
(c) 95 cm2
(d) 57 cm2



(b) 114 sq cm

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

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

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



$=8 \mathrm{~cm}$

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


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