Question Number 109508 by Study last updated on 24/Aug/20
Commented by som(math1967) last updated on 24/Aug/20
$$\:\frac{\mathrm{1}}{\mathrm{4}}\pi×\mathrm{2}^{\mathrm{2}} −\left\{\frac{\mathrm{1}}{\mathrm{2}}\pi×\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{1}×\mathrm{1}\right\} \\ $$$$\left(\pi−\frac{\pi}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}\right)\mathrm{cm}^{\mathrm{2}} \\ $$
Answered by Her_Majesty last updated on 24/Aug/20
$$\boldsymbol{{yes}}\:\boldsymbol{{I}}\:\boldsymbol{{can}} \\ $$
Answered by Rasheed.Sindhi last updated on 24/Aug/20
$$\left({i}\right){Quarter}\:{circle}\:{with}\:{radius}\:\mathrm{2}\:{cm} \\ $$$$\left({containing}\:{blue}\:\&\:{white}\:{part}\right) \\ $$$$\:\:\:\:\:\:\:\:=\frac{\pi\left(\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{4}}=\pi\:{cm}^{\mathrm{2}} \\ $$$${Radius}\:{of}\:{small}\:{circle}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{2}}\:/\mathrm{2} \\ $$$$\left({ii}\right){Small}\:{white}\:{cicle}-{part}\:{inside} \\ $$$${the}\:{square}= \\ $$$$\:\:\:\:{semi}-{circle}\:+{right}-{angled}\bigtriangleup \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\pi\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{1}×\mathrm{1} \\ $$$$\:\:\:=\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\pi+\mathrm{2}}{\mathrm{4}} \\ $$$${Blue}\:{Area}=\left({i}\right)−\left({ii}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\: \\ $$$$=\pi−\frac{\pi+\mathrm{2}}{\mathrm{4}}=\frac{\mathrm{4}\pi−\pi−\mathrm{2}}{\mathrm{4}}=\frac{\mathrm{3}\pi−\mathrm{2}}{\mathrm{4}} \\ $$
Commented by Rasheed.Sindhi last updated on 24/Aug/20
Commented by Rasheed.Sindhi last updated on 24/Aug/20
$${region}\mathrm{1}+{region}\mathrm{2}+{blue}\:{region} \\ $$$$\:\:\:\:={quarter}-{circle}\:{with}\:{radius}\:\mathrm{2} \\ $$
Commented by som(math1967) last updated on 24/Aug/20
$$\mathrm{Nice}\:\mathrm{sir} \\ $$
Commented by Rasheed.Sindhi last updated on 24/Aug/20
$$\:\:\:\:\:\:\:\mathcal{TH}{anks}\:\:{sir}! \\ $$