Question

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

Answer 1

(b) 114 sq cm

Using Pythagoras' theorem in $∆$ABC, we get:
$$\begin{gathered} A{C}^2=A{B}^2+B{C}^2 \\ \Rightarrow AB=\sqrt{A{C}^2-B{C}^2} \\ =\sqrt{{17}^2-{15}^2} \\ =8\mathrm{cm} \end{gathered}$$

$$\begin{gathered} \mathrm{Area}\mathrm{of}∆ABC=\frac{1}{2}\times AB\times BC \\ =\frac{1}{2}\times 8\times 15 \\ =60{\mathrm{cm}}^2 \end{gathered}$$

$$\begin{gathered} \mathrm{Area}\mathrm{of}∆BCD=\frac{1}{2}\times BD\times CD \\ =\frac{1}{2}\times 12\times 9 \\ =54{\mathrm{cm}}^2 \end{gathered}$$

∴ Area of quadrilateral ABCD$=\mathrm{Ar}\left(∆ABC\right)+\mathrm{Ar}\left(∆BCD\right)=54+60=114{\mathrm{cm}}^2$