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Question Number 77614 by BK last updated on 08/Jan/20

Commented by BK last updated on 08/Jan/20

$$\mathrm{find}\mathrm{the}\mathrm{radius}{\mathrm{A}}_{\mathrm{x}}$$

Commented by mr W last updated on 08/Jan/20

$$hablaespa\stackrel{\sim }{nol}?$$

Commented by mr W last updated on 08/Jan/20

$$t\stackrel{..}{urk}\underset{\varsigma }{ce}biliyormusun?$$

Commented by mr W last updated on 08/Jan/20

$$both?$$

Commented by mr W last updated on 08/Jan/20

$$ifnotboth,whichthen?$$

Commented by BK last updated on 08/Jan/20

$$\mathrm{im}\mathrm{from}\mathrm{turkey}$$

Answered by mr W last updated on 08/Jan/20

Commented by mr W last updated on 08/Jan/20

$$A_1=\frac{\pi {(a+c)}^2}{8}-\frac{\pi a^2}{8}-\frac{\pi c^2}{8}=\frac{\pi ca}{4}$$$$A_2=\frac{\pi ab}{4}$$$$A_2=\frac{\pi bc}{4}$$$$\frac{1}{A_1}+\frac{1}{A_2}+\frac{1}{A_3}=\frac{4}{\pi }(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca})=\frac{4(a+b+c)}{\pi abc}=\frac{1}{5}$$$$\Rightarrow \frac{abc}{a+b+c}=\frac{20}{\pi }$$$${\mathrm{\Delta }}_{ABC}=\sqrt{(a+b+c)abc}=(a+b+c)r$$$$\Rightarrow r=\sqrt{\frac{abc}{a+b+c}}=\sqrt{\frac{20}{\pi }}$$$$A_x=\pi r^2=20$$

Commented by BK last updated on 08/Jan/20

$$\mathrm{thanks}$$

Commented by Tawa11 last updated on 29/Dec/21

$$\mathrm{Great}\mathrm{sir}$$

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