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Question Number 28779 by ajfour last updated on 30/Jan/18

Commented by naka3546 last updated on 30/Jan/18

substitute x and y in a and b.

Answered by ajfour last updated on 30/Jan/18

$$Area=\frac{1}{2}{(a+b)}^2\cos \theta \sin \theta$$$$-\frac{1}{2}(a+b)a$$$$=\frac{1}{2}{(a+b)}^2\times \frac{ab}{a^2+b^2}-\frac{a(a+b)}{2}$$$$=\frac{a(a+b)}{2}\left[\frac{b(a+b)-(a^2+b^2)}{a^2+b^2}\right]$$$$\mathit{A}\mathit{r}\mathit{e}\mathit{a}=\frac{{\mathit{a}}^2({\mathit{b}}^2-{\mathit{a}}^2)}{2({\mathit{a}}^2+{\mathit{b}}^2)}.$$

Commented by naka3546 last updated on 02/Feb/18

$$thankyou,sir.$$

Answered by naka3546 last updated on 30/Jan/18

Answered by mrW2 last updated on 30/Jan/18

Commented by mrW2 last updated on 30/Jan/18

$$EC=\sqrt{a^2+b^2}$$$$\frac{AB}{a}=\frac{a+b}{b}=1+\frac{a}{b}$$$$\Rightarrow AB=a(1+\frac{a}{b})$$$$\frac{BD}{AB}=\frac{a}{\sqrt{a^2+b^2}}$$$$\Rightarrow BD=\frac{a^2}{\sqrt{a^2+b^2}}(1+\frac{a}{b})$$$$\frac{AD}{AB}=\frac{b}{\sqrt{a^2+b^2}}$$$$\Rightarrow AD=\frac{ab}{\sqrt{a^2+b^2}}(1+\frac{a}{b})$$$$\frac{BC}{a+b}=\frac{\sqrt{a^2+b^2}}{b}$$$$\Rightarrow BC=(1+\frac{a}{b})\sqrt{a^2+b^2}$$$$DE=(1+\frac{a}{b})\sqrt{a^2+b^2}-\sqrt{a^2+b^2}-\frac{a^2}{\sqrt{a^2+b^2}}(1+\frac{a}{b})$$$$=\frac{a(b-a)}{\sqrt{a^2+b^2}}$$$$$$$$A_{Blue}=\frac{1}{2}\times \frac{a(b-a)}{\sqrt{a^2+b^2}}\times \frac{ab}{\sqrt{a^2+b^2}}(1+\frac{a}{b})$$$$=\frac{a^2(b^2-a^2)}{2(a^2+b^2)}$$

Commented by ajfour last updated on 30/Jan/18

$$Thankyou,Sir.$$

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