Question Number 120133 by prakash jain last updated on 29/Oct/20
Commented by nimnim last updated on 29/Oct/20
$${Let}\:{the}\:{smallest}\:{diameter}\:{be}\:{x}\:{and}\:{the}\:{second}\:{be}\:{y} \\ $$$${then}\:{largest}\:{diameter}=\left({x}+{y}\right) \\ $$$${xy}=\mathrm{2}^{\mathrm{2}} =\mathrm{4} \\ $$$${Area}_{{shaded}} =\frac{\pi}{\mathrm{2}}\left(\frac{{x}+{y}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\pi}{\mathrm{2}}\left(\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\pi}{\mathrm{2}}\left(\frac{{y}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\pi}{\mathrm{8}}\left({x}^{\mathrm{2}} +\mathrm{2}{xy}+{y}^{\mathrm{2}} −{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\pi}{\mathrm{8}}\left(\mathrm{2}{xy}\right)=\frac{\pi}{\mathrm{8}}\left(\mathrm{8}\right)=\pi \\ $$
Commented by prakash jain last updated on 29/Oct/20
$$\mathrm{3}\:\mathrm{semi}-\mathrm{circles}\:\mathrm{and}\:\mathrm{length}\:\mathrm{of}\:\mathrm{line}\:\mathrm{are} \\ $$$$\mathrm{given}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{shaded}\:\mathrm{region} \\ $$
Commented by nimnim last updated on 29/Oct/20
$${I}\:{think}\:{its}\:\pi \\ $$
Commented by prakash jain last updated on 29/Oct/20
$$\mathrm{I}\:\mathrm{think}\:\mathrm{answer}\:\mathrm{is}\:\pi.\:\mathrm{But}\:\mathrm{need}\:\mathrm{to}\:\mathrm{prove} \\ $$$$\mathrm{that}\:\mathrm{area}\:\mathrm{stays}\:\mathrm{no}\:\mathrm{matter}\:\mathrm{where}\:\mathrm{the} \\ $$$$\mathrm{line}\:\mathrm{is}\:\mathrm{drawn} \\ $$
Commented by prakash jain last updated on 29/Oct/20
$$\mathrm{Thanks}. \\ $$