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Question Number 78216 by TawaTawa last updated on 15/Jan/20

Commented by TawaTawa last updated on 15/Jan/20

$$\mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{rectangle}.$$

Commented by mr W last updated on 15/Jan/20

$$radius=R$$$$DC=x$$$$areaofrectangle=xR$$$$\frac{x}{8}=\frac{8}{2R}$$$$\Rightarrow xR=\frac{8^2}{2}=areaofrectangle$$

Commented by Tony Lin last updated on 15/Jan/20

Commented by Tony Lin last updated on 15/Jan/20

$${(r-8cos\theta )}^2+{\left(8sin\theta \right)}^2=r^2$$$$rcos\theta =4$$$$Area=r\cdot 8cos\theta =32$$

Commented by TawaTawa last updated on 15/Jan/20

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by mr W last updated on 15/Jan/20

$$\frac{8^2}{2}=32$$

Commented by jagoll last updated on 15/Jan/20

$$verysimplesir$$

Commented by TawaTawa last updated on 15/Jan/20

$$\mathrm{Why}\mathrm{sir}.$$

Commented by TawaTawa last updated on 15/Jan/20

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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