Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 61424 by Tawa1 last updated on 02/Jun/19

Answered by mr W last updated on 02/Jun/19

Commented by mr W last updated on 02/Jun/19

$$\cos \alpha =\sin \beta =\frac{b}{2a}$$$$\Rightarrow \alpha ={\cos }^{-1}\frac{b}{2a}$$$$\Rightarrow \beta ={\sin }^{-1}\frac{b}{2a}$$$$A_{shade}=\pi b^2-\frac{b^2}{2}(2\alpha -\sin 2\alpha )-\frac{a^2}{2}(4\beta -\sin 4\beta )$$$$A_{shade}=\pi b^2-\frac{b^2}{2}[2{\cos }^{-1}\frac{b}{2a}-\sin \left(2{\cos }^{-1}\frac{b}{2a}\right)]-\frac{a^2}{2}[4{\sin }^{-1}\frac{b}{2a}-\sin \left(4{\sin }^{-1}\frac{b}{2a}\right)]$$

Commented by Tawa1 last updated on 02/Jun/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by alphaprime last updated on 02/Jun/19

Amazing

Commented by mr W last updated on 02/Jun/19

Commented by ajfour last updated on 02/Jun/19

$${shouldn}^{\prime }titbesimply$$$$A=\pi {\mathit{b}}^2-{\mathit{b}}^2\mathit{\alpha }-{\mathit{a}}^2(2\mathit{\beta }-\sin 2\mathit{\beta })$$

Commented by mr W last updated on 02/Jun/19

$$itcanbesimplifiedfurther.$$

Commented by mr W last updated on 02/Jun/19

Commented by mr W last updated on 02/Jun/19

$$A_{shade}=\pi b^2-A_1-A_2$$$$A_1=\frac{a^2}{2}(4\beta -\sin 4\beta )$$$$A_2=\frac{b^2}{2}(2\alpha -\sin 2\alpha )$$

Commented by mr W last updated on 02/Jun/19

$$A=\pi {\mathit{b}}^2-{\mathit{b}}^2\mathit{\alpha }-{\mathit{a}}^2(2\mathit{\beta }-\sin 2\mathit{\beta })$$$$iscertainlyabsolutelycorrect.$$

Commented by ajfour last updated on 02/Jun/19

Commented by Tawa1 last updated on 02/Jun/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by ajfour last updated on 02/Jun/19

$$thanks.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com