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Question-197771




Question Number 197771 by sonukgindia last updated on 28/Sep/23
Answered by som(math1967) last updated on 28/Sep/23
B=(1/2)×r^2 +2×((45)/(360))×πr^2   =(r^2 /2)+((πr^2 )/4)=(r^2 /2)(1+(π/2))squnit  A=(1/2)πr^2 −(r^2 /2)(1+(π/2))=(r^2 /2)((π/2)−1)squ  A:B=((π/2)−1):((π/2)+1)
$${B}=\frac{\mathrm{1}}{\mathrm{2}}×{r}^{\mathrm{2}} +\mathrm{2}×\frac{\mathrm{45}}{\mathrm{360}}×\pi{r}^{\mathrm{2}} \\ $$$$=\frac{{r}^{\mathrm{2}} }{\mathrm{2}}+\frac{\pi{r}^{\mathrm{2}} }{\mathrm{4}}=\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left(\mathrm{1}+\frac{\pi}{\mathrm{2}}\right){squnit} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2}}\pi{r}^{\mathrm{2}} −\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left(\mathrm{1}+\frac{\pi}{\mathrm{2}}\right)=\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}−\mathrm{1}\right){squ} \\ $$$${A}:{B}=\left(\frac{\pi}{\mathrm{2}}−\mathrm{1}\right):\left(\frac{\pi}{\mathrm{2}}+\mathrm{1}\right) \\ $$

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