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Question Number 160839 by amin96 last updated on 07/Dec/21

Commented by amin96 last updated on 07/Dec/21

yellow area=?

$$\boldsymbol{\mathrm{yellow}}\:\boldsymbol{\mathrm{area}}=? \\ $$

Answered by mr W last updated on 08/Dec/21

Commented by mr W last updated on 08/Dec/21

Method I (not so smart one) ((AB)/(sin (α+60)))=(2/(sin (α+30))) AB=((2 sin (α+60))/(sin (α+30))) ((DB)/(sin α))=(2/(sin (α+30))) DB=((2 sin α)/(sin (α+30))) S_1 =(1/2)×2×((2 sin (α+60))/(sin (α+30)))×sin 30=((sin (α+60))/(sin (α+30))) S_2 =(1/2)×2×((2 sin α)/(sin (α+30)))×sin 30=((sin α)/(sin (α+30))) S_(yellow) =((sin (α+60)+sin α)/(sin (α+30))) =((sin (α+30+30)+sin (α+30−30))/(sin (α+30))) =((2 sin (α+30) cos 30)/(sin (α+30))) =2 cos 30=(√3)

$$\boldsymbol{{Method}}\:\boldsymbol{{I}}\:\left({not}\:{so}\:{smart}\:{one}\right) \\ $$$$\frac{{AB}}{\mathrm{sin}\:\left(\alpha+\mathrm{60}\right)}=\frac{\mathrm{2}}{\mathrm{sin}\:\left(\alpha+\mathrm{30}\right)} \\ $$$${AB}=\frac{\mathrm{2}\:\mathrm{sin}\:\left(\alpha+\mathrm{60}\right)}{\mathrm{sin}\:\left(\alpha+\mathrm{30}\right)} \\ $$$$\frac{{DB}}{\mathrm{sin}\:\alpha}=\frac{\mathrm{2}}{\mathrm{sin}\:\left(\alpha+\mathrm{30}\right)} \\ $$$${DB}=\frac{\mathrm{2}\:\mathrm{sin}\:\alpha}{\mathrm{sin}\:\left(\alpha+\mathrm{30}\right)} \\ $$$${S}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}×\frac{\mathrm{2}\:\mathrm{sin}\:\left(\alpha+\mathrm{60}\right)}{\mathrm{sin}\:\left(\alpha+\mathrm{30}\right)}×\mathrm{sin}\:\mathrm{30}=\frac{\mathrm{sin}\:\left(\alpha+\mathrm{60}\right)}{\mathrm{sin}\:\left(\alpha+\mathrm{30}\right)} \\ $$$${S}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}×\frac{\mathrm{2}\:\mathrm{sin}\:\alpha}{\mathrm{sin}\:\left(\alpha+\mathrm{30}\right)}×\mathrm{sin}\:\mathrm{30}=\frac{\mathrm{sin}\:\alpha}{\mathrm{sin}\:\left(\alpha+\mathrm{30}\right)} \\ $$$${S}_{{yellow}} =\frac{\mathrm{sin}\:\left(\alpha+\mathrm{60}\right)+\mathrm{sin}\:\alpha}{\mathrm{sin}\:\left(\alpha+\mathrm{30}\right)} \\ $$$$=\frac{\mathrm{sin}\:\left(\alpha+\mathrm{30}+\mathrm{30}\right)+\mathrm{sin}\:\left(\alpha+\mathrm{30}−\mathrm{30}\right)}{\mathrm{sin}\:\left(\alpha+\mathrm{30}\right)} \\ $$$$=\frac{\mathrm{2}\:\mathrm{sin}\:\left(\alpha+\mathrm{30}\right)\:\mathrm{cos}\:\mathrm{30}}{\mathrm{sin}\:\left(\alpha+\mathrm{30}\right)} \\ $$$$=\mathrm{2}\:\mathrm{cos}\:\mathrm{30}=\sqrt{\mathrm{3}} \\ $$

Commented by Tawa11 last updated on 08/Dec/21

Great sir.

$$\mathrm{Great}\:\mathrm{sir}. \\ $$

Answered by mr W last updated on 08/Dec/21

Commented by mr W last updated on 08/Dec/21

Method II make ΔDEA=ΔCDB as shown yellow area=area of trapazoid ABDE DE//BA BD=AE BE=AD=2 area of trapazoid=((AD×BE×sin 60)/2) =2 sin 60=(√3)

$$\boldsymbol{{Method}}\:\boldsymbol{{II}} \\ $$$${make}\:\Delta{DEA}=\Delta{CDB}\:{as}\:{shown} \\ $$$${yellow}\:{area}={area}\:{of}\:{trapazoid}\:{ABDE} \\ $$$${DE}//{BA} \\ $$$${BD}={AE} \\ $$$${BE}={AD}=\mathrm{2} \\ $$$${area}\:{of}\:{trapazoid}=\frac{{AD}×{BE}×\mathrm{sin}\:\mathrm{60}}{\mathrm{2}} \\ $$$$=\mathrm{2}\:\mathrm{sin}\:\mathrm{60}=\sqrt{\mathrm{3}} \\ $$

Commented by Ari last updated on 09/Dec/21

Mr.Why area of trapezoid is (AD*BE*sin60)/2.This fomule is true for each trapezoid?

Commented by mr W last updated on 09/Dec/21

it′s true for any convex quadrilateral! A=((diagonal 1×diagonal 2×sin θ)/2)

$${it}'{s}\:{true}\:{for}\:{any}\:{convex}\:{quadrilateral}! \\ $$$${A}=\frac{{diagonal}\:\mathrm{1}×{diagonal}\:\mathrm{2}×\mathrm{sin}\:\theta}{\mathrm{2}} \\ $$

Commented by Ari last updated on 09/Dec/21

ok,thanks!

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