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 3808 by Rasheed Soomro last updated on 21/Dec/15

$$Achorddividesthecircleintwo$$$$segments,havingareas{s}_{1}and{s}_2.$$$$Ifdiameter,perpendiculartothis$$$$chordiscutinto1:3bythechord,whatis{s}_1:s_2?$$$$$$

Commented by Rasheed Soomro last updated on 21/Dec/15

$$LetABisachordandCD\left(\mathrm{\perp }AB\right)isadiameter.$$$$LetABcutsCDatpointO$$$$IfCO:OD=1:3,whatis{s}_1:s_2$$$$s_{1}and{s}_{2}areareasoftwosegmentsofthe$$$$circleproducedbychordAB.$$

Answered by Yozzii last updated on 21/Dec/15

$$(IntegralCalculuscanalternatively$$$$beusedtofindthevalueof\frac{s_1}{s_2}.)$$$$Drawacirclewithdiameterlengthdand$$$$centreQ.Adiameterofthecirclenormally$$$$cutsachordABonthecirclesuchthat$$$$dispartitionedintheratio1:3.Let$$$$LFbesuchadiameterandDbethe$$$$pointofintersectionofthechordAB$$$$anddiameterLF.Wethushave$$$$\mid LD\mid =3aand\mid DF\mid =a(a>0)sothatDF:LD=1:3.$$$$Therefore,d=3a+a=4a\Rightarrow radiusr=2a.$$$$Completetheisoscelestriangle△AQB$$$$andlet\mathrm{\angle }AQB=\theta .Thearea{A}_{1}ofthesectorAB$$$$ofthecircleisthen$$$$A_1=\frac{1}{2}{r}^2\theta =\frac{1}{2}\times 4a^2\theta =2a^2\theta .$$$$Theareaof△AQB,denotedby{A}_2,isgivenby$$$$A_2=\frac{1}{2}2a\times 2a\times sin\theta =2a^{2}sin\theta .$$$$Now,let{s}_{1}betheareabetweenthechord$$$$ABandthearcAB.$$$$\Rightarrow s_1=A_1-A_2=2a^2(\theta -sin\theta )$$$$Hence,theareaoftheothersegment$$$$ofthecircleis{s}_2=\pi r^2-s_1.$$$$s_2=4\pi a^2-2a^2(\theta -sin\theta )$$$$s_2=2a^2(2\pi +sin\theta -\theta )$$$$Therefore,\frac{s_1}{s_2}=\frac{\theta -sin\theta }{2\pi +sin\theta -\theta }.$$$$Returningto△AQB,wecanyield$$$$aright-angledtriangle△AQD$$$$with\mathrm{\angle }AQD=\frac{1}{2}\mathrm{\angle }AQB=\frac{\theta }{2}$$$$and\mathrm{\angle }QDA=\frac{\pi }{2}.$$$$Wethenhavecos\frac{\theta }{2}=\frac{QD}{AQ}=\frac{a}{2a}=\frac{1}{2}$$$$\Rightarrow \frac{\theta }{2}=\frac{\pi }{3}\Rightarrow \theta =\frac{2\pi }{3}.$$$$\therefore \frac{s_1}{s_2}=\frac{\frac{2\pi }{3}-sin\frac{2\pi }{3}}{2\pi +sin\frac{2\pi }{3}-\frac{2\pi }{3}}=\frac{\frac{2\pi }{3}-\frac{\sqrt{3}}{2}}{2\pi +\frac{\sqrt{3}}{2}-\frac{2\pi }{3}}$$$$\frac{s_1}{s_2}=\frac{4\pi -3\sqrt{3}}{8\pi +3\sqrt{3}}$$$$$$$$$$$$$$$$$$$$$$$$$$

Commented by Rasheed Soomro last updated on 21/Dec/15

$${}^{″}....Adiameterofthecirclenormally$$$$cutsachordAB\underset{-}{onthecircle}suchthat$$$$dispartitionedintheratio1:3...{.}^{″}$$$$Whatdoyoumeanbythis?$$

Commented by Yozzii last updated on 21/Dec/15

$$normally\Rightarrow perpendicularly\left(inmathterms\right)$$

Commented by Yozzii last updated on 21/Dec/15

$$$$$$Thediametercutsthechordat90°$$$$withthechorddivindingthelengthof$$$$thediameterintheratio1:3.$$

Commented by Yozzii last updated on 21/Dec/15

$$Thequestion{hadn}^{\prime }tspecifiedwhether$$$$s_1>s_{2}or{s}_2>s_{1}sotruly\frac{s_1}{s_2}cantake$$$$twovaluesthatarereciprocals$$$$ofeachother.$$

Commented by Rasheed Soomro last updated on 22/Dec/15

$$Thanks!Misunderstandinginunderstanding$$$$language.Igot{}^{\prime }{\mathit{n}\mathit{o}\mathit{r}\mathit{m}\mathit{a}\mathit{l}\mathit{l}\mathit{y}}^{\prime }innormallanguage,$$$$notasmathematicalterm.$$

Commented by Rasheed Soomro last updated on 21/Dec/15

$${\mathcal{G}}^{{\mathcal{O}}^V\mathcal{O}}\mathcal{D}\mathcal{A}pproach!$$

Answered by Yozzii last updated on 21/Dec/15

$$(Integralcalculusmethodbasedon$$$$initialinformationfromfirstanswer.)$$$$$$$$DefineacircleCwithequation$$$$x^2+y^2=4a^2,a>0.$$$$ForC,let{s}_{1}betheareabetweenthe$$$$chordABandthearcAB.Wecanlet$$$$thediameternormaltoABbeconcurrent$$$$withthex-axis.$$$$\therefore s_1={\int }_a^{2a}ydx$$$$s_1={\int }_a^{2a}\sqrt{4a^2-x^2}dx$$$$Letx=2asin\theta \Rightarrow dx=2acos\theta d\theta$$$$and\sqrt{4a^2-x^2}=2acos\theta .$$$$Atx=a\Rightarrow sin\theta =1/2\Rightarrow \theta =\frac{\pi }{6}.$$$$Atx=2a\Rightarrow sin\theta =1\Rightarrow \theta =\pi /2.$$$$\therefore s_1=2{\int }_{\pi /6}^{\pi /2}4a^2{cos}^2\theta d\theta$$$$s_1=4a^2{\int }_{\pi /6}^{\pi /2}(1+cos2\theta )d\theta$$$$s_1=4a^2(\theta +\frac{1}{2}sin2\theta ){\mid }_{\pi /6}^{\pi /2}$$$$s_1=4a^2(\frac{\pi }{2}-\frac{\pi }{6}+\frac{1}{2}sin\pi -\frac{1}{2}sin\frac{\pi }{3})$$$$s_1=4a^2(\frac{\pi }{3}-\frac{\sqrt{3}}{4})$$$$s_1=\frac{a^2}{3}(4\pi -3\sqrt{3})$$$$Thearea{s}_{2}oftheothersegmentis$$$$givenby{s}_2=\pi r^2-s_1.$$$$\therefore s_2=4\pi a^2-4a^2\frac{4\pi -3\sqrt{3}}{12}$$$$s_2=4a^2(\pi -\frac{4\pi -3\sqrt{3}}{12})$$$$s_2=4a^2\left(\frac{12\pi -4\pi +3\sqrt{3}}{12}\right)$$$$s_2=\frac{a^2}{3}(8\pi +3\sqrt{3})$$$$\therefore \frac{s_1}{s_2}=\frac{4\pi -3\sqrt{3}}{8\pi +3\sqrt{3}}$$$$$$

Commented by Rasheed Soomro last updated on 21/Dec/15

$$\mathbb{A}\mathrm{n}\mathbb{O}\mathbf{ther}$$$$\stackrel{{\mathcal{O}}^{\mathbb{V}}\mathcal{O}}{\mathcal{G}\mathcal{D}}$$$$\mathrm{Approach}!$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com