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Question Number 15558 by ajfour last updated on 11/Jun/17

Commented by ajfour last updated on 11/Jun/17

$$\mathbf{Find}\mathbf{the}\mathbf{areas}\mathbf{I},\mathbf{II},\mathbf{III},\mathbf{and}\mathbf{IV}$$$$\mathbf{in}\mathbf{terms}\mathbf{of}\mathbf{a},\mathbf{and}\mathbf{b}.$$$$\mathbf{Curve}\mathbf{APD}\mathbf{is}\mathbf{a}\mathbf{semicircle}.\mathbf{ABCD}$$$$\mathbf{is}\mathbf{a}\mathbf{rectangle}\mathbf{with}\mathbf{side}\mathbf{a}>\mathbf{b}/2.$$

Answered by mrW1 last updated on 11/Jun/17

$$\alpha ={\tan }^{-1}\frac{\mathrm{a}}{\mathrm{b}}$$$$\mathrm{\angle }\mathrm{DOP}=\pi -2\alpha$$$$$$$${\mathrm{A}}_{\mathrm{III}}=\frac{1}{2}{\left(\frac{\mathrm{b}}{2}\right)}^2[\pi -2\alpha -\sin (\pi -2\alpha )]$$$$=\frac{{\mathrm{b}}^2}{8}[\pi -2\alpha -\sin 2\alpha ]$$$$=\frac{{\mathrm{b}}^2}{8}[\pi -2\alpha -2\times \frac{\mathrm{ab}}{{\mathrm{a}}^2+{\mathrm{b}}^2}]$$$$=\frac{{\mathrm{b}}^2}{4}[\frac{\pi }{2}-{\tan }^{-1}\frac{\mathrm{a}}{\mathrm{b}}-\frac{\mathrm{ab}}{{\mathrm{a}}^2+{\mathrm{b}}^2}]$$$$$$$${\mathrm{A}}_{\mathrm{IV}}=\frac{\pi }{2}{\left(\frac{\mathrm{b}}{2}\right)}^2-{\mathrm{A}}_{\mathrm{III}}$$$$=\frac{{\mathrm{b}}^2}{4}[{\tan }^{-1}\frac{\mathrm{a}}{\mathrm{b}}+\frac{\mathrm{ab}}{{\mathrm{a}}^2+{\mathrm{b}}^2}]$$$$$$$${\mathrm{A}}_{\mathrm{II}}=\frac{\mathrm{ab}}{2}-{\mathrm{A}}_{\mathrm{III}}$$$$=\frac{\mathrm{ab}}{2}-\frac{{\mathrm{b}}^2}{4}[\frac{\pi }{2}-{\tan }^{-1}\frac{\mathrm{a}}{\mathrm{b}}-\frac{\mathrm{ab}}{{\mathrm{a}}^2+{\mathrm{b}}^2}]$$$$$$$${\mathrm{A}}_{\mathrm{I}}=\frac{\mathrm{ab}}{2}-{\mathrm{A}}_{\mathrm{IV}}$$$$=\frac{\mathrm{ab}}{2}-\frac{{\mathrm{b}}^2}{4}[{\tan }^{-1}\frac{\mathrm{a}}{\mathrm{b}}+\frac{\mathrm{ab}}{{\mathrm{a}}^2+{\mathrm{b}}^2}]$$

Commented by mrW1 last updated on 11/Jun/17

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