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Question Number 86789 by TawaTawa1 last updated on 31/Mar/20

Commented by jagoll last updated on 31/Mar/20

Answered by jagoll last updated on 31/Mar/20

Commented by jagoll last updated on 31/Mar/20

$$\mathrm{miss}\mathrm{tawa}.\mathrm{i}\mathrm{think}\mathrm{you}\mathrm{can}\mathrm{solve}\mathrm{it}$$

Commented by TawaTawa1 last updated on 31/Mar/20

$$\mathrm{I}\mathrm{appreciate}\mathrm{your}\mathrm{time}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}$$

Commented by TawaTawa1 last updated on 31/Mar/20

$$\mathrm{I}\mathrm{did}\mathrm{not}\mathrm{understand}\mathrm{the}\mathrm{first}\mathrm{step}.$$

Commented by jagoll last updated on 31/Mar/20

$$\mathrm{i}\mathrm{use}\mathrm{descartes}\mathrm{theorem}$$

Commented by TawaTawa1 last updated on 31/Mar/20

$$\mathrm{But}\mathrm{sir},\mathrm{you}\mathrm{have}\mathrm{not}\mathrm{find}\mathrm{the}\mathrm{area}.$$$$\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{area}\mathrm{sir}\mathrm{please}.$$$$\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{shaded}$$

Commented by jagoll last updated on 31/Mar/20

$$\mathrm{the}\mathrm{area}\mathrm{can}\mathrm{already}\mathrm{be}\mathrm{calculated}$$$$\mathrm{by}\mathrm{substracting}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}$$$$\mathrm{triangle}\mathrm{with}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{circle}$$

Commented by TawaTawa1 last updated on 31/Mar/20

$$\mathrm{The}\mathrm{three}\mathrm{circle}?$$

Commented by TawaTawa1 last updated on 01/Apr/20

$$\mathrm{Sir}\mathrm{jagoll}\mathrm{please}\mathrm{complete}\mathrm{solution}.\mathrm{Please}\mathrm{sir}.$$

Commented by jagoll last updated on 01/Apr/20

$$\mathrm{my}\mathrm{answer}\mathrm{same}\mathrm{with}\mathrm{mr}\mathrm{w}$$

Answered by mr W last updated on 31/Mar/20

$${\left(\frac{5}{2}\right)}^2+{(5-r)}^2={(\frac{5}{2}+r)}^2$$$$\Rightarrow r=\frac{5}{3}$$$$\tan \theta =\frac{5-\frac{5}{3}}{\frac{5}{2}}=\frac{4}{3}$$$$A_{shaded}=\frac{5(5-\frac{5}{3})}{2}-2\times \frac{1}{2}{\left(\frac{5}{2}\right)}^2\theta -\frac{1}{2}{\left(\frac{5}{3}\right)}^2(\pi -2\theta )$$$$=\frac{25}{3}-\frac{25}{4}\theta -\frac{1}{2}\times \frac{25}{9}(\pi -2\theta )$$$$=\frac{25}{36}(12-2\pi -5\theta )$$$$=\frac{25}{36}(12-2\pi -5{\tan }^{-1}\frac{4}{3})$$

Commented by TawaTawa1 last updated on 31/Mar/20

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

Commented by TawaTawa1 last updated on 01/Apr/20

$$\mathrm{Sir}\mathrm{i}\mathrm{got}-180.51\mathrm{when}\mathrm{i}\mathrm{press}\mathrm{calculator}.$$$$\mathrm{Can}\mathrm{the}\mathrm{area}\mathrm{be}\mathrm{negative}\mathrm{or}\mathrm{there}\mathrm{is}\mathrm{a}\mathrm{mistake}\mathrm{somewhere}.$$

Commented by mr W last updated on 01/Apr/20

$$sorry.i{can}^{\prime }tteachyouhowtouse$$$$calculator.itseemsyou{don}^{\prime }tmake$$$$differencebetweenradanddegree,$$$$ijustguess.$$

Commented by john santu last updated on 01/Apr/20

$$=0.750235$$

Commented by TawaTawa1 last updated on 01/Apr/20

$$\mathbf{Sir}\mathbf{did}\mathbf{you}\mathbf{use}\pi =180\mathrm{degree}.$$

Commented by john santu last updated on 01/Apr/20

Commented by TawaTawa1 last updated on 01/Apr/20

$$\mathrm{Ohh}.\mathrm{You}\mathrm{work}\mathrm{in}\mathrm{radian}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}.$$

Commented by TawaTawa1 last updated on 01/Apr/20

$$\mathrm{Am}\mathrm{sorry}\mathrm{sir}\mathrm{mrW}.\mathrm{I}\mathrm{press}\mathrm{i}\mathrm{work}\mathrm{in}\mathrm{degree}.$$

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