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Question Number 79254 by mr W last updated on 23/Jan/20

Commented by mr W last updated on 23/Jan/20

$$Ifthesidelengthofthesquareis10,$$$$find$$$$1)theradiusofthesmallestcircle$$$$2)theshadedarea$$

Commented by key of knowledge last updated on 23/Jan/20

$$1)\mathrm{r}=\frac{5}{3}2)\mathrm{i}\mathrm{think}\mathrm{need}(\int )!$$

Answered by john santu last updated on 24/Jan/20

Commented by mr W last updated on 24/Jan/20

$$notcorrectsir!$$$$itseemsbutitisnot:APnot\mathrm{\perp }PB,AP\ne 5.$$

Commented by john santu last updated on 24/Jan/20

$$r_A+r=5...\left(i\right)$$$${(r_A+5)}^2=25+{(10-r_A)}^2$$$$r_A^2+10r_A+25=25+100-20r_A+r_A^2$$$$30r_A=100\Rightarrow r_A=\frac{10}{3}$$$$nowwework\left(i\right)r=5-\frac{10}{3}=\frac{5}{3}$$$$$$

Commented by john santu last updated on 24/Jan/20

$$proveitif{it}^{\prime }snotrightmister$$

Commented by mr W last updated on 24/Jan/20

$$sir:whatimeantaboveis$$$$itseemsbutitisnotproved:AP\mathrm{\perp }PB,AP=5.$$$$youjustassumedthesearetrue.$$$$infacti{can}^{\prime }tseeinyoursolution$$$$thatyouutilisedtheconditionthat$$$$thesmallestcirclealsotangentsthe$$$$largestquartocircle.$$$$ihavenotcalculatedthisquestion,$$$$soihavenoresultyet.maybeyour$$$$resultisregardlesscorrect.then$$$${it}^{\prime }sbyaccident,ithink.$$$$i^{\prime }llpostmysolutionandcompare.$$

Commented by mr W last updated on 24/Jan/20

$$ihavecalculated.yourresult\left(values\right)$$$$iscorrect.$$

Answered by mr W last updated on 24/Jan/20

Commented by mr W last updated on 24/Jan/20

$$\left(PartI\right)$$$$2R=10\Rightarrow R=5$$$$OB=2R-s$$$$AB=R+s$$$${(R+s)}^2=R^2+{(2R-s)}^2$$$$3s=2R$$$$\Rightarrow s=\frac{2R}{3}=\frac{10}{3}$$$$OC=2R-r$$$$BC=s+r$$$$let\beta =\mathrm{\angle }BOC$$$$\cos \beta =\frac{{(2R-s)}^2+{(2R-r)}^2-{(s+r)}^2}{2(2R-s)(2R-r)}$$$$\Rightarrow \cos \beta =\frac{2(R-r)}{2R-r}$$$$AC=R+r$$$$let\alpha =\mathrm{\angle }AOC$$$$\cos \alpha =\frac{R^2+{(2R-r)}^2-{(R+r)}^2}{2R(2R-r)}$$$$\Rightarrow \cos \alpha =\frac{2R-3r}{2R-r}=\sin \beta$$$${\cos }^2\beta +{\sin }^2\beta =1$$$$\frac{4{(R-r)}^2+{(2R-3r)}^2}{{(2R-r)}^2}=1$$$$4R^2-8Rr+4r^2+4R^2-12Rr+9r^2=4R^2-4Rr+r^2$$$$R^2-4Rr+3r^2=0$$$$(R-r)(R-3r)=0$$$$\Rightarrow r=R\Rightarrow notthesolution$$$$\Rightarrow r=\frac{R}{3}=\frac{5}{3}$$

Commented by john santu last updated on 24/Jan/20

$${(10-r)}^2=5^2+{(5+r)}^2$$$$100-20r+r^2=25+25+10r+r^2$$$$50=30r\Rightarrow r=\frac{5}{3}$$

Commented by mr W last updated on 24/Jan/20

$${(10-r)}^2=5^2+{(5+r)}^{2}meansOA\mathrm{\perp }AC,$$$$butithinkthisisnotyetproved.or$$$$howdidyougetthatOA\mathrm{\perp }AC?$$

Commented by john santu last updated on 24/Jan/20

$$mister.wecannotclaimour$$$$wayisright,sowedeemdifferent$$$$wayswrong.itisshortsighted.$$$${everyone}^{\prime }spointofviewis$$$$different$$

Commented by john santu last updated on 24/Jan/20

Commented by john santu last updated on 24/Jan/20

$$looksir$$

Commented by mr W last updated on 24/Jan/20

$$discussionisalwaysgood.$$$$btwyourdiagram{didn}^{\prime }tanswermy$$$$questionhowtogetthatOA\mathrm{\perp }AC.it$$$$saysonlythatAC\mathrm{\perp }thetangentline.$$

Answered by mr W last updated on 24/Jan/20

Commented by mr W last updated on 24/Jan/20

$$\left(PartII\right)$$$$ascalculatedinpartIwehavegot$$$$s=\frac{2R}{3}=\frac{10}{3}$$$$r=\frac{R}{3}=\frac{5}{3}$$$$BC=s+r=\frac{2R}{3}+\frac{R}{3}=R=OA$$$$OB=2R-s=2R-\frac{2R}{3}=\frac{4R}{3}$$$$AC=R+r=R+\frac{R}{3}=\frac{4R}{3}=OB$$$$\Rightarrow OACBisrectangle.$$$$\tan \beta =\frac{R}{\frac{4R}{3}}=\frac{3}{4}$$$$\Rightarrow \beta ={\tan }^{-1}\frac{3}{4}$$$$\gamma =\frac{\pi }{2}+\beta$$$$A_{\stackrel{⌢}{ODE}}=\frac{\beta {\left(2R\right)}^2}{2}=2R^2\beta$$$$A_{\mathrm{\Delta }OBC}=\frac{1}{2}\times R\times \frac{4R}{3}=\frac{2R^2}{3}$$$$A_{\stackrel{⌢}{BDF}}=\frac{\pi }{4}{\left(\frac{2R}{3}\right)}^2=\frac{\pi R^2}{9}$$$$A_{\stackrel{⌢}{FCE}}=\frac{\gamma }{2}{\left(\frac{R}{3}\right)}^2=\frac{R^2}{18}(\frac{\pi }{2}+\beta )$$$$A_{shaded}=A_{\stackrel{⌢}{ODE}}-A_{\mathrm{\Delta }OBC}-A_{\stackrel{⌢}{BDF}}-A_{\stackrel{⌢}{FCE}}$$$$=2R^2\beta -\frac{2R^2}{3}-\frac{\pi R^2}{9}-\frac{R^2}{18}(\frac{\pi }{2}+\beta )$$$$=\frac{(70\beta -5\pi -24)R^2}{36}$$$$\Rightarrow A_{shaded}=\frac{(70{\tan }^{-1}\frac{3}{4}-5\pi -24)R^2}{36}\approx 0.148253R^2$$$$\approx 3.706329$$

Answered by mr W last updated on 24/Jan/20

Commented by mr W last updated on 24/Jan/20

$$hereisanotherwaytosolveusing$$$$the{Descartes}^{\prime }theorem$$$$\left(seeQ77681formoredetails\right)$$$$$$$$R_1=2R=10$$$$R_2=R=5$$$$R_3=s=?$$$$R_4=r=?$$$$$$$$applying{Descartes}^{\prime }theoremon$$$$R_1,2\times R_{2}and{R}_3:$$$${(-\frac{1}{R_1}+\frac{2}{R_2}+\frac{1}{R_3})}^2=2(\frac{1}{R_1^2}+\frac{2}{R_2^2}+\frac{1}{R_3^2})$$$${(-\frac{1}{2R}+\frac{2}{R}+\frac{1}{s})}^2=2(\frac{1}{4R^2}+\frac{2}{R^2}+\frac{1}{s^2})$$$$4\frac{R^2}{s^2}-12\frac{R}{s}+9=0$$$${(2\frac{R}{s}-3)}^2=0$$$$\Rightarrow s=\frac{2R}{3}=\frac{10}{3}$$$$$$$$applying{Descartes}^{\prime }theoremon$$$$R_1,R_2,R_{3}and{R}_4:$$$${(-\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4})}^2=2(\frac{1}{R_1^2}+\frac{1}{R_2^2}+\frac{1}{R_3^2}+\frac{1}{R_4^2})$$$${(-\frac{1}{2R}+\frac{1}{R}+\frac{3}{2R}+\frac{1}{r})}^2=2(\frac{1}{4R^2}+\frac{1}{R^2}+\frac{9}{4R^2}+\frac{1}{r^2})$$$$\frac{R^2}{r^2}-4\frac{R}{r}+3=0$$$$(\frac{R}{r}-1)(\frac{R}{r}-3)=0$$$$\Rightarrow r=R\Rightarrow notthesearchedcircle$$$$\Rightarrow r=\frac{R}{3}=\frac{5}{3}$$

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