Question-86453

Question Number 86453 by TawaTawa1 last updated on 28/Mar/20

Commented by john santu last updated on 28/Mar/20

2a = distance the centre point of two circle?

$$\mathrm{2a}\:=\:\mathrm{distance}\:\mathrm{the}\:\mathrm{centre}\:\mathrm{point}\:\mathrm{of} \\ $$$$\mathrm{two}\:\mathrm{circle}? \\ $$

Commented by TawaTawa1 last updated on 28/Mar/20

$$\mathrm{Yes}\:\mathrm{sir} \\ $$

Commented by jagoll last updated on 29/Mar/20

$$\mathrm{what}\:\mathrm{this}\:\mathrm{answer}\:\mathrm{miss}\:\mathrm{tawa}? \\ $$

Answered by jagoll last updated on 29/Mar/20

Commented by jagoll last updated on 29/Mar/20

cos θ = (a/R) ⇒θ = cos^(−1) ((a/R)) the area of 1 shaded area is = ((2cos^(−1) ((a/R)))/(2π))×πR^2 − (1/2)R^2 sin 2θ = R^2 cos^(−1) ((a/R)) − R^2 (((a (√(R^2 −a^2 )))/R^2 ) ) = R^2 cos^(−1) ((a/R))−a (√(R^2 −a^2 )) now the total area your request = 2R^2 cos^(−1) ((a/R))−2a (√(R^2 −a^2 ))

$$\mathrm{cos}\:\theta\:=\:\frac{\mathrm{a}}{\mathrm{R}}\:\Rightarrow\theta\:=\:\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{a}}{\mathrm{R}}\right) \\ $$$$\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{1}\:\mathrm{shaded}\:\mathrm{area}\:\mathrm{is}\: \\ $$$$=\:\frac{\mathrm{2cos}^{−\mathrm{1}} \left(\frac{\mathrm{a}}{\mathrm{R}}\right)}{\mathrm{2}\pi}×\pi\mathrm{R}^{\mathrm{2}} −\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{R}^{\mathrm{2}} \:\mathrm{sin}\:\mathrm{2}\theta \\ $$$$=\:\mathrm{R}^{\mathrm{2}} \:\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{a}}{\mathrm{R}}\right)\:−\:\mathrm{R}^{\mathrm{2}} \:\left(\frac{\mathrm{a}\:\sqrt{\mathrm{R}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }}{\mathrm{R}^{\mathrm{2}} }\:\right) \\ $$$$=\:\mathrm{R}^{\mathrm{2}} \:\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{a}}{\mathrm{R}}\right)−\mathrm{a}\:\sqrt{\mathrm{R}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} } \\ $$$$\mathrm{now}\:\mathrm{the}\:\mathrm{total}\:\mathrm{area}\:\mathrm{your}\:\mathrm{request} \\ $$$$=\:\mathrm{2R}^{\mathrm{2}} \:\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{a}}{\mathrm{R}}\right)−\mathrm{2a}\:\sqrt{\mathrm{R}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} } \\ $$

Commented by TawaTawa1 last updated on 29/Mar/20

God bless you sir. I appreciate. I will use your answer. i don′t know the answer.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}.\:\mathrm{I}\:\mathrm{will}\:\mathrm{use}\:\mathrm{your}\:\mathrm{answer}. \\ $$$$\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{the}\:\mathrm{answer}. \\ $$

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