Question

In the given figure, ABCD is a rectangle in which diag. AC = 17 cm, ∠BCA = θ and $\sin \mathrm{\theta }=\frac{8}{17}$.
Find (i) the area of rect. ABCD, (ii) the perimeter of rect. ABCD.

Answer 1

$$\begin{gathered} \mathrm{Given}:\mathrm{In}∆ABC, \\ AC=17\mathrm{cm} \\ \sin \theta =\frac{8}{17} \\ \mathrm{Since},\sin \theta =\frac{P}{H} \\ \Rightarrow P=8\mathrm{and}H=17 \\ \mathrm{Using}\mathrm{Pythagoras}\mathrm{theorem}, \\ {P}^2+B^2=H^2 \\ \Rightarrow 8^2+B^2={17}^2 \\ \Rightarrow B^2=289-64 \\ \Rightarrow B^2=225 \\ \Rightarrow B=15 \\ \mathrm{Therefore}, \\ AB=8\mathrm{cm}\mathrm{and}BC=15\mathrm{cm} \\ \mathrm{Therefore}, \\ \left(\mathrm{i}\right)\mathrm{Area}\mathrm{of}\mathrm{rectangle}ABCD=AB\times BC \\ =8\times 15 \\ =120{\mathrm{cm}}^2 \\ \left(ii\right)\mathrm{Perimeter}\mathrm{of}\mathrm{rectangle}ABCD=2\left(AB+BC\right) \\ =2\left(8+15\right) \\ =2\left(23\right) \\ =46\mathrm{cm} \end{gathered}$$