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Question-154065




Question Number 154065 by peter frank last updated on 13/Sep/21
Answered by Rasheed.Sindhi last updated on 14/Sep/21
Area of the badge=Two semi-circles                                         +One rectangle  2(((πx^2 )/2))+xy=20  πx^2 +xy=20  y=((20−πx^2 )/x)  Perimeter=Two semi-circles                                 +2lengths+2widths  p=2πx+2y+2x     =2πx+2(((20−πx^2 )/x))+2x     =((2πx^2 +40−2πx^2 +2x^2 )/x)     =2x+((40)/x)
$${Area}\:{of}\:{the}\:{badge}={Two}\:{semi}-{circles} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{One}\:{rectangle} \\ $$$$\mathrm{2}\left(\frac{\pi{x}^{\mathrm{2}} }{\mathrm{2}}\right)+{xy}=\mathrm{20} \\ $$$$\pi{x}^{\mathrm{2}} +{xy}=\mathrm{20} \\ $$$${y}=\frac{\mathrm{20}−\pi{x}^{\mathrm{2}} }{{x}} \\ $$$${Perimeter}={Two}\:{semi}-{circles} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{2}{lengths}+\mathrm{2}{widths} \\ $$$${p}=\mathrm{2}\pi{x}+\mathrm{2}{y}+\mathrm{2}{x} \\ $$$$\:\:\:=\mathrm{2}\pi{x}+\mathrm{2}\left(\frac{\mathrm{20}−\pi{x}^{\mathrm{2}} }{{x}}\right)+\mathrm{2}{x} \\ $$$$\:\:\:=\frac{\cancel{\mathrm{2}\pi{x}^{\mathrm{2}} }+\mathrm{40}−\cancel{\mathrm{2}\pi{x}^{\mathrm{2}} }+\mathrm{2}{x}^{\mathrm{2}} }{{x}} \\ $$$$\:\:\:=\mathrm{2}{x}+\frac{\mathrm{40}}{{x}} \\ $$
Commented by peter frank last updated on 14/Sep/21
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$

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