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Question Number 121553 by I want to learn more last updated on 09/Nov/20

Commented by I want to learn more last updated on 09/Nov/20

$$\mathrm{Area}\mathrm{of}\mathrm{shaded}$$

Answered by TANMAY PANACEA last updated on 09/Nov/20

$$LetA(0,0)B(2,0)D(0,1)C(2,1)$$$$eqnAB=xaxisAD=yaxis$$$$centresmallcircle(1,k)radius=r$$$$k+r=1$$$${(1-0)}^2+{(k-0)}^2={(1+r)}^2[distancebetween$$$$centreAandcentreofsmallcircle]$$$$1+{(1-r)}^2=1+2r+r^2$$$$1+1-2r+r^2=1+2r+r^2$$$$4r=1\to r=\frac{1}{4}$$$$\mathit{s}\mathit{h}\mathit{a}\mathit{d}\mathit{e}\mathit{d}\mathit{a}\mathit{r}\mathit{e}\mathit{a}.$$$$2\times 1-2\times \frac{1}{4}\times \pi 1^2-\pi \times {\left(\frac{1}{4}\right)}^2$$$$=2-\frac{\pi }{2}-\frac{\pi }{16}=$$$$=\frac{32-8\pi -\pi }{16}=\frac{32-9\pi }{16}$$$$$$

Commented by I want to learn more last updated on 09/Nov/20

$$\mathrm{Thanks}\mathrm{sir}.$$

Commented by TANMAY PANACEA last updated on 09/Nov/20

$$mostwelcome$$

Answered by mr W last updated on 09/Nov/20

$$R=1$$$${(R+r)}^2=R^2+{(R-r)}^2$$$$4r=R$$$$r=\frac{R}{4}$$$$shaded=2R\times R-\frac{\pi R^2}{2}-\pi r^2$$$$=\left(\frac{32-9\pi }{16}\right)R^2=\frac{32-9\pi }{16}=0.233$$

Commented by I want to learn more last updated on 10/Nov/20

$$\mathrm{I}\mathrm{appreciate}\mathrm{sir}$$

Commented by I want to learn more last updated on 10/Nov/20

$$\mathrm{Sir},\mathrm{help}\mathrm{me}\mathrm{solve}\mathrm{the}\mathrm{permutation}\mathrm{Q}121552$$

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