Question and Answers Forum

All Questions      Topic List

Mensuration Questions

Previous in All Question      Next in All Question      

Previous in Mensuration      Next in Mensuration      

Question Number 187253 by Rupesh123 last updated on 15/Feb/23

Answered by MikeH last updated on 15/Feb/23

$$\left(\mathrm{i}\right)\mathrm{Area}\mathrm{of}△BCD=\frac{1}{2}\left(4\right)\left(4\right)\sin x=8\sin x$$$$\mathrm{Area}\mathrm{of}\mathrm{\Delta }ABD=\frac{1}{2}\left(3\right)\left(4\right)\sin (90-x)=6\cos x$$$$\mathrm{total}\mathrm{area}=\mathrm{area}\mathrm{of}\mathrm{\Delta }BCD+\mathrm{area}\mathrm{of}\mathrm{\Delta }ABD$$$$\Rightarrow A=6\cos x+8\sin x◼$$

Commented by Rupesh123 last updated on 15/Feb/23

Good job!

Answered by MikeH last updated on 15/Feb/23

$$\left(\mathrm{ii}\right)A=R\cos (x-\alpha )$$$$\Rightarrow 6\cos x+8\sin x\equiv R\cos x\cos \alpha +R\sin x\sin \alpha$$$$\Rightarrow R\cos \alpha =6.....\left(1\right)$$$$R\sin \alpha =8.....\left(2\right)$$$${\left(1\right)}^2+{\left(2\right)}^2$$$$\Rightarrow R=\sqrt{6^2+8^2}=10$$$$\left(2\right)/\left(1\right)\Rightarrow \tan \alpha =\frac{4}{3}\Rightarrow \alpha =53.13°$$$$\overline{)\begin{array}{c}A=10\cos (x-53.13°)\end{array}}$$

Answered by MikeH last updated on 15/Feb/23

$$\left(\mathrm{iii}\right)\mathrm{maximum}\mathrm{area}\mathrm{occurs}\mathrm{when}\mathrm{cosine}\mathrm{is}$$$$\mathrm{maximum}$$$$\Rightarrow maxA\mathrm{rea}=10{\mathrm{cm}}^2$$

Answered by MikeH last updated on 15/Feb/23

$$\left(\mathrm{iv}\right)A=10\cos (x-53.13°)$$$$A=7$$$$\Rightarrow \cos (x-53.13°)=\frac{7}{10}$$$$\Rightarrow x-53.13°=360°n\pm 45.57$$$$\Rightarrow x=360°n\pm 45.57+53.13$$$$0°⩽x<90°\Rightarrow n=0\Rightarrow x=7.56°$$

Commented by Rupesh123 last updated on 15/Feb/23

Good my friend!

Terms of Service

Privacy Policy

Contact: info@tinkutara.com