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Question Number 67254 by TawaTawa last updated on 24/Aug/19

Commented by Rasheed.Sindhi last updated on 24/Aug/19

$$AnalyticalMethod$$$$x-axis:HorizanalChord$$$$y-axis:VerticalChord$$$$Threepointsonthecircle:$$$$A(6,0),B(-2,0)\mathrm{\&}C(0,-3)$$$$FindoutcircumcentreO(x,y)of△ABC$$$$Radius=\mid OA\mid or\mid OB\mid or\mid OC\mid$$$$$$

Commented by MJS last updated on 24/Aug/19

$$\mathrm{Sir}{\mathrm{Rasheed}}^{\prime }\mathrm{s}\mathrm{idea}\mathrm{completed}$$$$a=8b=\sqrt{2^2+3^2}=\sqrt{13}c=\sqrt{3^2+6^2}=\sqrt{45}$$$$\delta =\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}$$$$R=\frac{abc}{\delta }=\frac{\sqrt{65}}{2}[{\mathrm{Heron}}^{\prime }\mathrm{s}\mathrm{formula};\mathrm{search}\mathrm{the}\mathrm{web}\mathrm{for}\mathrm{it}]$$

Commented by TawaTawa last updated on 25/Aug/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sirs}.$$

Commented by TawaTawa last updated on 25/Aug/19

$$\mathrm{Sir},\mathrm{please}\mathrm{label}\mathrm{the}\mathrm{diagram}\mathrm{to}\mathrm{back}\mathrm{up}\mathrm{the}\mathrm{solution}.\mathrm{Thanks}\mathrm{sir}.$$

Commented by Rasheed.Sindhi last updated on 25/Aug/19

$${\mathcal{T}}^{{\mathcal{H}}^{\mathcal{A}}\mathcal{N}}\mathcal{K}s\mathcal{S}ır\mathrm{MJS}!$$

Commented by TawaTawa last updated on 25/Aug/19

$$\mathrm{Sir},\mathrm{how}\mathrm{did}\mathrm{you}\mathrm{get}\mathrm{a},\mathrm{b},\mathrm{c}\mathrm{please}$$

Commented by Tony Lin last updated on 25/Aug/19

Commented by Tony Lin last updated on 25/Aug/19

$$R=\frac{abc}{4\times △Area}$$$$=\frac{8\times \sqrt{13}\times 3\sqrt{5}}{4\times \left(\frac{1}{2}\times 8\times 3\right)}$$$$=\frac{\sqrt{65}}{2}$$

Commented by Tony Lin last updated on 25/Aug/19

$$youcanremember\frac{abc}{4\times △Area}thisformula$$$$\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}=2R$$$$\Rightarrow \frac{abc}{\frac{1}{2}bcsinA}=\frac{abc}{\frac{1}{2}acsinB}=\frac{abc}{\frac{1}{2}absinC}=4R$$$$\Rightarrow \frac{abc}{4\times △Area}=R$$

Commented by TawaTawa last updated on 25/Aug/19

$$\mathrm{I}\mathrm{appreciate}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{Now},\mathrm{i}\mathrm{understand}$$

Answered by mr W last updated on 24/Aug/19

Commented by mr W last updated on 24/Aug/19

$$\frac{h}{2}=\frac{6}{3}$$$$\Rightarrow h=4$$$$R^2-{\left(\frac{2+6}{2}\right)}^2={(\frac{3+4}{2}-3)}^2$$$$R^2=16+\frac{1}{4}=\frac{65}{4}$$$$\Rightarrow R=\frac{\sqrt{65}}{2}$$

Commented by TawaTawa last updated on 25/Aug/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir},\mathrm{i}\mathrm{get}\mathrm{the}\mathrm{rest}\mathrm{but}\mathrm{i}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{understand}\mathrm{how}\mathrm{you}\mathrm{used}$$$${\mathrm{R}}^2-{\left(\frac{2+6}{2}\right)}^2={(\frac{3+4}{2}-3)}^2.$$$$\mathrm{R}\mathrm{is}\mathrm{not}\mathrm{labelled}.\mathrm{please}\mathrm{help}\mathrm{sir}.$$

Commented by mr W last updated on 25/Aug/19

Commented by TawaTawa last updated on 25/Aug/19

$$\mathrm{Sir},\mathrm{please}\mathrm{give}\mathrm{me}\mathrm{any}\mathrm{question}\mathrm{on}\mathrm{finding}\mathrm{area}\mathrm{of}\mathrm{shaded}\mathrm{portion}\mathrm{i}\mathrm{should}$$$$\mathrm{try}.\mathrm{Thanks}\mathrm{sir}.$$

Commented by TawaTawa last updated on 25/Aug/19

$$\mathrm{Wow},\mathrm{i}\mathrm{understand}\mathrm{it}\mathrm{now}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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