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Question Number 183393 by cherokeesay last updated on 25/Dec/22

Answered by mr W last updated on 25/Dec/22

Commented by mr W last updated on 25/Dec/22

$$\frac{x}{2}=\frac{2}{2+2}$$$$\Rightarrow x=1$$$$R=\frac{1+2+2}{2}=\frac{5}{2}$$$$\cos \theta =\frac{{(R-r)}^2+{(R-1)}^2-{(r+1)}^2}{2(R-r)(R-1)}$$$$\cos \theta =\frac{{(r+2)}^2-{(R-1)}^2-{(R-r)}^2}{2(R-r)(R-1)}$$$${(r+2)}^2-{(R-1)}^2-{(R-r)}^2={(R-r)}^2+{(R-1)}^2-{(r+1)}^2$$$$9(r-\frac{1}{4})-\frac{9}{4}=7(\frac{3}{4}-r)+\frac{9}{4}$$$$\Rightarrow r=\frac{3}{4}$$$$S=\frac{\pi \times 2^2}{4}=\pi$$$$S_1=\pi r^2$$$$S_2=\frac{\pi \times 1^2}{2}=\frac{\pi }{2}$$$$\frac{S}{S_1+S_2}=\frac{\pi }{\pi r^2+\frac{\pi }{2}}=\frac{2}{1+2r^2}=\frac{16}{17}✓$$

Commented by mr W last updated on 25/Dec/22

$$errorisfixed.thanks!$$

Commented by cherokeesay last updated on 25/Dec/22

$$c^{\prime }est\frac{16}{17},ilyaerreurdansvoscalculssir.$$

Answered by Acem last updated on 25/Dec/22

Commented by Acem last updated on 27/Dec/22

$$BDisahalfofthechordso\mid BD{\mid }^2=\mid AD\mid .\mid DC\mid$$

Commented by Acem last updated on 25/Dec/22

$$Continue...$$

Commented by Acem last updated on 26/Dec/22

Commented by Acem last updated on 27/Dec/22

Answered by HeferH last updated on 25/Dec/22

Commented by cherokeesay last updated on 25/Dec/22

$$ilyaerreurdansvoscalculs!$$$$c^{\prime }est\frac{16}{17}$$

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