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Question Number 208639 by Tawa11 last updated on 20/Jun/24

Commented by mr W last updated on 20/Jun/24

$$isitaredsquare?$$$$doesonecornerofthesquarelie$$$$onthecenterofsemicircle?$$

Commented by Tawa11 last updated on 20/Jun/24

$$\mathrm{Yes}\mathrm{sir}.$$

Commented by som(math1967) last updated on 20/Jun/24

$$igot4\times \sqrt{\frac{5}{3}}$$

Answered by AlagaIbile last updated on 20/Jun/24

$$\mathit{x}=\sqrt{\frac{80}{3}}\mathit{c}\mathit{m}=4\sqrt{\frac{5}{3}}\mathit{c}\mathit{m}$$

Commented by Tawa11 last updated on 20/Jun/24

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

Answered by mr W last updated on 20/Jun/24

$$sidelengthofsquarea=\sqrt{5}$$$$radiusofsemicircler$$$$a^2=(r-a)(r+a)$$$$\Rightarrow r^2=2a^2$$$$\frac{x}{2r}=\frac{r}{\sqrt{r^2+a^2}}$$$$\Rightarrow x=\frac{2r^2}{\sqrt{r^2+a^2}}=\frac{4a}{\sqrt{3}}=\frac{4\sqrt{5}}{\sqrt{3}}=\frac{4\sqrt{15}}{3}✓$$

Commented by Tawa11 last updated on 20/Jun/24

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

Commented by mr W last updated on 20/Jun/24

Commented by Tawa11 last updated on 20/Jun/24

$$\mathrm{Wow},\mathrm{I}\mathrm{understand}\mathrm{now}\mathrm{sir}.$$

Answered by A5T last updated on 20/Jun/24

Commented by A5T last updated on 20/Jun/24

$$s\times s=5\Rightarrow s=\sqrt{5};s^2+s^2=r^2\Rightarrow r=\sqrt{10}$$$$\frac{x-k}{2r}=\frac{r}{x}\Rightarrow 2r^2=x^2-kx\Rightarrow 20=x(x-k)...\left(i\right)$$$${(x-k)}^2=s^2+r^2=15\Rightarrow x-k=\sqrt{15}...\left(ii\right)$$$$\left(i\right)\mathrm{\&}\left(ii\right)\Rightarrow x=\frac{20}{\sqrt{15}}=\frac{4\sqrt{15}}{3}$$

Commented by Tawa11 last updated on 20/Jun/24

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

Commented by Tawa11 last updated on 20/Jun/24

$$\mathrm{Sir},\mathrm{am}\mathrm{confused}\mathrm{on}.$$$$\frac{\mathrm{x}-\mathrm{k}}{2\mathrm{r}}=\frac{\mathrm{r}}{\mathrm{x}}$$$$\mathrm{Please}\mathrm{explain}\mathrm{the}\mathrm{rule}\mathrm{that}\mathrm{gives}\mathrm{us}\mathrm{that}\mathrm{sir}.$$

Commented by A5T last updated on 20/Jun/24

Commented by A5T last updated on 20/Jun/24

$$△DFCissimilartoABC$$$$\Rightarrow \frac{DC=x-k}{AC=2r}=\frac{FC=r}{BC=x}\Rightarrow 2r^2=x(x-k)$$

Commented by Tawa11 last updated on 20/Jun/24

$$\mathrm{Wow},\mathrm{I}\mathrm{understand}\mathrm{now}\mathrm{sir}$$

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