Question Number 219119 by Spillover last updated on 19/Apr/25
Answered by A5T last updated on 20/Apr/25
![R=6 Area=4×[2(((6^2 π)/4)−(6^2 /2))]=8[((36π−72)/4)]=72(π−2)](https://www.tinkutara.com/question/Q219128.png)
$$\mathrm{R}=\mathrm{6} \\ $$$$\mathrm{Area}=\mathrm{4}×\left[\mathrm{2}\left(\frac{\mathrm{6}^{\mathrm{2}} \pi}{\mathrm{4}}−\frac{\mathrm{6}^{\mathrm{2}} }{\mathrm{2}}\right)\right]=\mathrm{8}\left[\frac{\mathrm{36}\pi−\mathrm{72}}{\mathrm{4}}\right]=\mathrm{72}\left(\pi−\mathrm{2}\right) \\ $$
Answered by Nicholas666 last updated on 20/Apr/25
$$\mathrm{36}\left(\mathrm{6}−\pi\right){cm}^{\mathrm{2}} \\ $$
Commented by Nicholas666 last updated on 20/Apr/25
$${r}\:=\:\frac{\mathrm{12}\:{cm}}{\mathrm{2}}\:=\:\mathrm{6}\:{cm}\: \\ $$$${A}_{{Sequare}\:} \:=\:\mathrm{12}^{\mathrm{2}} \:=\:\mathrm{144}{cm}^{\mathrm{2}} \\ $$$${A}_{{unshade}_{−} {segment}\:} =\:\frac{\mathrm{1}}{\mathrm{4}}\pi\left(\mathrm{6}^{\mathrm{2}} \right)−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{6}\right)\left(\mathrm{6}\right)\:=\:\mathrm{9}\pi−\mathrm{18}{cm}^{\mathrm{2}} \:\: \\ $$$${A}_{{unshaded}_{−} {total}} =\mathrm{4}\left(\mathrm{9}\pi−\mathrm{18}\right)=\mathrm{36}\pi−\mathrm{72}{cm}^{\mathrm{2}} \:\:\: \\ $$$${A}_{{shaded}} =\:{A}_{{sequare}} −{A}_{{unshaded}_{−} {total}\:} = \\ $$$$\Rightarrow\mathrm{144}\:−\:\left(\mathrm{36}\pi\:−\mathrm{72}\right)=\:\mathrm{144}−\mathrm{36}\pi+\mathrm{72}= \\ $$$$\Rightarrow\mathrm{216}\:−\mathrm{36}\pi{cm}\: \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by Spillover last updated on 20/Apr/25
Answered by Spillover last updated on 20/Apr/25
