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Question Number 159864 by cherokeesay last updated on 21/Nov/21

Answered by som(math1967) last updated on 22/Nov/21

((MP)/(AM))=sin30⇒AM=2R BL=KN=(√(MN^2 −MK^2 )) =(√((R+r)^2 +(R−r)^2 ))=(√(4rR))=2(√R) ((LC)/(LN))=cot30⇒LC=(√3)LN=(√3)r=(√3) ((AB)/(BC))=tan60 ((2R+R)/(2(√R)+(√3)))=tan60 3R=2(√(3R))+3 3(R−1)=2(√(3R)) 9(R−1)^2 =12R 3R^2 +3−6R=4R 3R^2 −10R+3=0 R=3 ∴AB=3×3=9 BC=ABtan30=3(√3) AC=(√(27+81))=6(√3) perimeter=(6(√3)+3(√3)+9)unit

$$\frac{{MP}}{{AM}}={sin}\mathrm{30}\Rightarrow{AM}=\mathrm{2}{R} \\ $$$${BL}={KN}=\sqrt{{MN}^{\mathrm{2}} −{MK}^{\mathrm{2}} } \\ $$$$=\sqrt{\left({R}+{r}\right)^{\mathrm{2}} +\left({R}−{r}\right)^{\mathrm{2}} }=\sqrt{\mathrm{4}{rR}}=\mathrm{2}\sqrt{{R}} \\ $$$$\frac{{LC}}{{LN}}={cot}\mathrm{30}\Rightarrow{LC}=\sqrt{\mathrm{3}}{LN}=\sqrt{\mathrm{3}}{r}=\sqrt{\mathrm{3}} \\ $$$$\frac{{AB}}{{BC}}={tan}\mathrm{60} \\ $$$$\frac{\mathrm{2}{R}+{R}}{\mathrm{2}\sqrt{{R}}+\sqrt{\mathrm{3}}}={tan}\mathrm{60} \\ $$$$\mathrm{3}{R}=\mathrm{2}\sqrt{\mathrm{3}{R}}+\mathrm{3} \\ $$$$\mathrm{3}\left({R}−\mathrm{1}\right)=\mathrm{2}\sqrt{\mathrm{3}{R}} \\ $$$$\mathrm{9}\left({R}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{12}{R} \\ $$$$\mathrm{3}{R}^{\mathrm{2}} +\mathrm{3}−\mathrm{6}{R}=\mathrm{4}{R} \\ $$$$\mathrm{3}{R}^{\mathrm{2}} −\mathrm{10}{R}+\mathrm{3}=\mathrm{0} \\ $$$${R}=\mathrm{3} \\ $$$$\therefore{AB}=\mathrm{3}×\mathrm{3}=\mathrm{9} \\ $$$${BC}={ABtan}\mathrm{30}=\mathrm{3}\sqrt{\mathrm{3}} \\ $$$${AC}=\sqrt{\mathrm{27}+\mathrm{81}}=\mathrm{6}\sqrt{\mathrm{3}} \\ $$$${perimeter}=\left(\mathrm{6}\sqrt{\mathrm{3}}+\mathrm{3}\sqrt{\mathrm{3}}+\mathrm{9}\right){unit} \\ $$

Commented by som(math1967) last updated on 22/Nov/21

Commented by cherokeesay last updated on 22/Nov/21

thank you sir !

$${thank}\:{you}\:{sir}\:! \\ $$

Answered by mr W last updated on 22/Nov/21

Commented by mr W last updated on 22/Nov/21

((EF)/(EG))=sin 30° ((R−r)/(R+r))=(1/2) ⇒R=3r=3 AD=DC=BC=(R/(tan 30°))=(√3)R=3(√3) AE=EC=(R/(sin 30°))=2R=6 perimeter P=3+3×3(√3)+6=9(1+(√3))

$$\frac{{EF}}{{EG}}=\mathrm{sin}\:\mathrm{30}° \\ $$$$\frac{{R}−{r}}{{R}+{r}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{R}=\mathrm{3}{r}=\mathrm{3} \\ $$$${AD}={DC}={BC}=\frac{{R}}{\mathrm{tan}\:\mathrm{30}°}=\sqrt{\mathrm{3}}{R}=\mathrm{3}\sqrt{\mathrm{3}} \\ $$$${AE}={EC}=\frac{{R}}{\mathrm{sin}\:\mathrm{30}°}=\mathrm{2}{R}=\mathrm{6} \\ $$$${perimeter}\:{P}=\mathrm{3}+\mathrm{3}×\mathrm{3}\sqrt{\mathrm{3}}+\mathrm{6}=\mathrm{9}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right) \\ $$

Commented by som(math1967) last updated on 22/Nov/21

Good way sir

$${Good}\:{way}\:{sir} \\ $$

Commented by cherokeesay last updated on 22/Nov/21

very nice ! thank you !

$${very}\:{nice}\:! \\ $$$${thank}\:{you}\:! \\ $$

Commented by Tawa11 last updated on 22/Nov/21

Great

$$\mathrm{Great} \\ $$

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