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Question Number 94448 by mr W last updated on 18/May/20

Commented by behi83417@gmail.com last updated on 18/May/20

Commented by behi83417@gmail.com last updated on 19/May/20

$$\mathrm{A}\stackrel{▴}{\mathrm{D}C},\mathrm{is}\mathrm{isoscale}.\mathrm{and}\mathrm{so}\mathrm{on}:\mathrm{A}\stackrel{▴}{\mathrm{B}D},\mathrm{A}\stackrel{▴}{\mathrm{B}C},\mathrm{B}\stackrel{▴}{\mathrm{D}C}.$$$$\mathrm{so}:\mathrm{\angle }\mathrm{A}=\mathrm{\angle }\mathrm{B}=\mathrm{\angle }\mathrm{C}=\mathrm{\angle }\mathrm{D}=\frac{1}{4}\times 360={90}^{\circ }$$$$\Rightarrow \mathrm{ABCD}:\mathrm{square}(\Rightarrow \mathrm{blue}=\mathrm{square})$$$$\mathrm{\angle }{\mathrm{B}}_1={\mathrm{tg}}^{-1}\left(\frac{1}{3}\right),\mathrm{\angle }{\mathrm{B}}_2={\mathrm{tg}}^{-1}\left(\frac{2}{3}\right)$$$$\mathrm{x}=\sin \left({\mathrm{tg}}^{-1}\left(\frac{1}{3}\right)\right)=\frac{1}{\sqrt{10}},\mathrm{y}=\frac{3}{\sqrt{10}}$$$$\mathrm{z}=\sqrt{10}-\frac{1}{\sqrt{10}}-\frac{3}{\sqrt{10}}=\frac{6}{\sqrt{10}}\left(\mathrm{blue}\mathrm{side}\right)$$$$\Rightarrow {\mathrm{S}}_{\mathrm{blue}}={\mathrm{z}}^2={\left(\frac{6}{\sqrt{10}}\right)}^2=\frac{36}{10}=\frac{18}{5}$$

Answered by MJS last updated on 18/May/20

$$x^2+y^2=1$$$${(y+z)}^2+y^2=9$$$$3^2+1^2={(x+y+z)}^2$$$$\mathrm{blue}\mathrm{square}=z^2$$$$\mathrm{I}\mathrm{get}{z}^2=\frac{18}{5}$$

Answered by john santu last updated on 19/May/20

$$\frac{\mathrm{blue}\mathrm{area}}{\mathrm{square}\mathrm{area}}=\frac{2}{5}$$$$\mathrm{then}\mathrm{blue}\mathrm{area}=\frac{2}{5}\times 9=\frac{18}{5}$$

Commented by mr W last updated on 19/May/20

$$thankyouallfordifferentways!$$

Answered by mr W last updated on 19/May/20

Commented by mr W last updated on 19/May/20

$$shadedtrianglesarecongruent.$$$$s=\sqrt{10}-\frac{1}{\sqrt{10}}-\frac{3}{\sqrt{10}}=\frac{3\sqrt{10}}{5}$$$$bluearea=s^2=\frac{18}{5}$$

Commented by mr W last updated on 19/May/20

$$or$$$$areaofashadedbigtriangle=\frac{3}{2}$$$$areaofashadedsmalltriangle=\frac{3}{2}\times \frac{1}{10}$$$$bluearea=square-4\times bigtri.+4\times smalltri.$$$$=3\times 3-4\times \frac{3}{2}+4\times \frac{3}{2}\times \frac{1}{10}=\frac{18}{5}$$

Commented by PRITHWISH SEN 2 last updated on 19/May/20

$$\mathrm{Actually}\mathrm{I}\mathrm{am}\mathrm{waiting}\mathrm{for}\mathrm{something}\mathrm{different}$$$$\mathrm{like}\mathrm{this}.$$

Commented by mr W last updated on 19/May/20

$$pleaseshowusthissomethingdifferent$$$$sir!$$

Commented by PRITHWISH SEN 2 last updated on 19/May/20

$$\tan \theta =\frac{1}{3}\Rightarrow \cos \theta =\frac{3}{\sqrt{10}}$$$$\mathrm{AP}=\cos \theta ,\mathrm{AQ}=3\cos \theta \therefore \mathrm{PQ}=\mathrm{AQ}-\mathrm{AP}=2\cos \theta$$$$\therefore \mathrm{PQ}=\frac{6}{\sqrt{10}}\therefore \mathrm{Area}=\frac{36}{10}=\frac{18}{5}$$$$\mathrm{this}\mathrm{is}\mathrm{my}\mathrm{try}\mathrm{sir}$$

Commented by PRITHWISH SEN 2 last updated on 19/May/20

Commented by PRITHWISH SEN 2 last updated on 19/May/20

$$\mathrm{another}\mathrm{approach}$$$$\mathrm{let}\mathrm{the}\mathrm{square}\mathrm{ABCD}\mathrm{has}\mathrm{been}\mathrm{twist}\mathrm{for}\mathrm{an}\mathrm{angle}$$$$\mathrm{of}\theta \mathrm{and}\mathrm{contracted}\mathrm{by}\mathrm{a}\mathrm{factor}\mathrm{of}\mathrm{K}$$$$\mathrm{now}\mathrm{K}=\frac{\mathrm{AB}}{\mathrm{AG}}=\frac{3}{\sqrt{10}}<1$$$$\mathrm{then}\mathrm{PQ}=\mathrm{K}.\mathrm{FB}=\frac{6}{\sqrt{10}}$$$$\mathrm{then}\mathrm{Area}=\frac{36}{10}=\frac{18}{5}$$

Commented by PRITHWISH SEN 2 last updated on 19/May/20

$$\mathrm{Let}\mathrm{B}\mathrm{be}\mathrm{a}\mathrm{complex}\mathrm{number}\mathrm{on}\mathrm{the}\mathrm{complex}\mathrm{plane}$$$$\mathrm{with}\mathrm{AB}\mathrm{as}\mathrm{a}\mathrm{real}\mathrm{axis}\mathrm{and}\mathrm{A}\mathrm{as}\mathrm{the}\mathrm{origin}$$$$\mathrm{then}\mathrm{B}={3\mathrm{e}}^{\mathbf{i}0}$$$$\mathrm{now}\mathrm{a}\mathrm{complex}\mathrm{number}\mathrm{is}\mathrm{multiplied}\mathrm{which}\mathrm{effect}$$$$\mathrm{a}\mathrm{contraction}$$$$\mathrm{then}\mathrm{let}\mathrm{the}\mathrm{number}\mathrm{be}{\mathrm{re}}^{\mathbf{i}\mathit{\theta }}$$$$\therefore \mathrm{Q}={3\mathrm{re}}^{\mathbf{i}\mathit{\theta }}\Rightarrow \mid \mathrm{AQ}\mid =3\mathrm{r}=\frac{9}{\sqrt{10}}$$$$\mathrm{then}\mathrm{r}=\frac{3}{\sqrt{10}}\mathrm{now}\mathrm{if}\mathrm{another}\mathrm{complex}\mathrm{number}$$$${2\mathrm{e}}^{\mathbf{i}0}\mathrm{be}\mathrm{multiplied}\mathrm{by}\mathrm{the}\mathrm{same}\mathrm{complex}\mathrm{number}$$$$\mathrm{then}\mathrm{it}\mathrm{turns}\mathrm{to}={2\mathrm{re}}^{\mathbf{i}\mathit{\theta }}$$$$\mathrm{now}\mid {2\mathrm{re}}^{\mathbf{i}\mathit{\theta }}\mid =\mid \mathrm{PQ}\mid =\frac{6}{\sqrt{10}}$$$$\therefore \mathrm{the}\mathrm{Area}={\left(\mathrm{PQ}\right)}^2=\frac{18}{5}$$

Commented by mr W last updated on 19/May/20

$$great!$$

Commented by PRITHWISH SEN 2 last updated on 19/May/20

$$\mathrm{No}\mathrm{sir},\mathrm{yours}\mathrm{is}\mathrm{beautiful}\mathrm{as}\mathrm{it}\mathrm{is}\mathrm{simple}.\mathrm{I}\mathrm{wish}$$$$\mathrm{one}\mathrm{day}\mathrm{I}\mathrm{will}\mathrm{able}\mathrm{to}\mathrm{think}\mathrm{like}\mathrm{you}.\mathrm{Thank}\mathrm{you}$$$$\mathrm{sir}.$$

Commented by mr W last updated on 19/May/20

$$sometimes{it}^{\prime }sanicefeelingto$$$$searchfororeventofindthat$$$$somethingdifferent.$$

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