Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 27627 by Rasheed.Sindhi last updated on 11/Jan/18

Commented by Rasheed.Sindhi last updated on 11/Jan/18

$$\mathrm{Given}\mathrm{a}\mathrm{circle}\mathrm{of}\mathrm{radius}\mathrm{r}.\mathrm{AB}\mathrm{\&}\mathrm{CD}$$$$\mathrm{are}\mathrm{two}\mathrm{chords}\mathrm{intersecting}\mathrm{at}\mathrm{E}$$$$\mathrm{OX}\mathrm{and}\mathrm{OY}\mathrm{are}\mathrm{respective}\mathrm{distances}$$$$\mathrm{of}\mathrm{chords}\mathrm{CD}\mathrm{and}\mathrm{AB}\mathrm{from}\mathrm{center}$$$$\mathrm{of}\mathrm{the}\mathrm{circle}.$$$$\mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{region}\mathrm{AEC}$$$$(\mathrm{The}\mathrm{region}\mathrm{bound}\mathrm{by}\mathrm{line}\mathrm{segments}$$$$\mathrm{AE},\mathrm{CE}\mathrm{and}\mathrm{the}\mathrm{minor}\mathrm{arc}\stackrel{⌢}{\mathrm{AC}})\mathrm{when}$$$$\mathrm{OX}=\mathrm{x},\mathrm{OY}=\mathrm{y}$$$$\mathrm{\angle }\mathrm{XOY}=\theta$$

Answered by mrW2 last updated on 14/Jan/18

Commented by mrW2 last updated on 14/Jan/18

$$Eqn.ofcircle:$$$$x^2+y^2=r^2$$$$\Rightarrow y=\sqrt{r^2-x^2}$$$$$$$$Eqn.ofCD:$$$$y=p+kxwithk=-\frac{1}{\tan \theta }$$$$CDpassesthroughF,$$$$b\sin \theta =p-\frac{1}{\tan \theta }\times b\cos \theta$$$$\Rightarrow p=b(\sin \theta +\frac{{\cos }^2\theta }{\sin \theta })=\frac{b}{\sin \theta }$$$$\Rightarrow y=\frac{b}{\sin \theta }-\frac{1}{\tan \theta }x=\frac{b-x\cos \theta }{\sin \theta }$$$$$$$$PointD(x_D,y_D):$$$$y=y_D=\sqrt{r^2-x^2}=\frac{b-x\cos \theta }{\sin \theta }$$$${\sin }^2\theta (r^2-x^2)=b^2-2b\cos \theta x+{\cos }^2\theta x^2)$$$$x^2-2b\cos \theta x-(r^2{\sin }^2\theta -b^2)=0$$$$\Rightarrow x=x_D=\frac{2b\cos \theta -\sqrt{4b^4{\cos }^2\theta +4(r^2{\sin }^2\theta -b^2)}}{2}$$$$\Rightarrow x_D=b[\cos \theta -\sin \theta \sqrt{{\left(\frac{r}{b}\right)}^2-1}]$$$$$$$$x_C=b[\cos \theta +\sin \theta \sqrt{{\left(\frac{r}{b}\right)}^2-1}]⩾a(!)$$$$$$$$AreaA={\int }_{x_D}^a(\sqrt{r^2-x^2}-\frac{b}{\sin \theta }+\frac{1}{\tan \theta }x)dx$$$$={[\frac{r^2}{2}{\sin }^{-1}\frac{x}{r}+\frac{x\sqrt{r^2-x^2}}{2}-\frac{bx}{\sin \theta }+\frac{x^2}{2\tan \theta }]}_{x_D}^a$$$$\Rightarrow A=[\frac{r^2}{2}({\sin }^{-1}\frac{a}{r}-{\sin }^{-1}\frac{x_D}{r})+\frac{a\sqrt{r^2-a^2}-x_D\sqrt{r^2-x_D^2}}{2}-\frac{b(a-x_D)}{\sin \theta }+\frac{a^2-x_D^2}{2\tan \theta }]$$

Commented by Rasheed.Sindhi last updated on 14/Jan/18

$$⅁\mathrm{r}\mathrm{\exists }\mathrm{\forall }+\mathbb{S}\mathrm{ir}!$$$$\underset{\mathcal{SSS}..\mathcal{S}}{\mathrm{Than}\mathbb{k}}\vee \mathrm{ery}\mathrm{m}⋓⋐\nparallel !$$$$\mathrm{Could}\mathrm{you}\mathrm{do}\mathrm{also}\mathrm{using}\mathrm{non}-$$$$\mathrm{analytic}\mathrm{approach}.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com