Question Number 108331 by I want to learn more last updated on 16/Aug/20
Answered by mr W last updated on 16/Aug/20
Commented by mr W last updated on 16/Aug/20
$${R}={a}+{b} \\ $$$$\frac{\mathrm{2}{a}}{{BD}}=\frac{{BD}}{\mathrm{2}{b}} \\ $$$$\Rightarrow\mathrm{4}{ab}={BD}^{\mathrm{2}} =\mathrm{2}^{\mathrm{2}} =\mathrm{4} \\ $$$$\Rightarrow{ab}=\mathrm{1} \\ $$$${A}_{{shaded}} =\frac{\pi}{\mathrm{2}}\left({R}^{\mathrm{2}} −{a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)=\pi{ab}=\pi \\ $$
Commented by I want to learn more last updated on 16/Aug/20
$$\mathrm{I}\:\mathrm{appreciate},\:\mathrm{thanks}\:\mathrm{sir}. \\ $$
Commented by I want to learn more last updated on 16/Aug/20
$$\mathrm{Sir},\:\mathrm{answer}\:\mathrm{in}\:\mathrm{book}\:\mathrm{is}\:\:\mathrm{4}\pi,\:\:\:\mathrm{That}\:\mathrm{means}\:\mathrm{the}\:\mathrm{textbook}\:\mathrm{is}\:\mathrm{a}\:\mathrm{mistake}. \\ $$
Commented by mr W last updated on 16/Aug/20
$${yes}. \\ $$
Commented by I want to learn more last updated on 16/Aug/20
$$\mathrm{Thanks}\:\mathrm{sir}\:\mathrm{for}\:\mathrm{your}\:\mathrm{time}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$
Commented by mr W last updated on 16/Aug/20
$${just}\:{look}\:{at}\:{the}\:{special}\:{case}: \\ $$$${BD}={R}=\mathrm{2} \\ $$$$\Rightarrow{a}={b}=\frac{{R}}{\mathrm{2}}=\mathrm{1} \\ $$$${A}_{{shade}} =\frac{\pi}{\mathrm{2}}\left(\mathrm{2}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} \right)=\pi \\ $$
Commented by I want to learn more last updated on 17/Aug/20
$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{more}\:\mathrm{clarification}\:\mathrm{sir} \\ $$