Question-195611

Question Number 195611 by Mingma last updated on 05/Aug/23

Answered by mr W last updated on 06/Aug/23

a=side length of small pentagon b=side length of big pentagon b=2a sin 54° a=2r sin 36° ⇒r=(a/(2 sin 36°)) b=2R sin 36° ⇒R=(b/(2 sin 36°))=((a sin 54°)/(sin 36°)) total area =((5 R^2 sin 72°)/2)+((a^2 sin 72°)/2) shaded area=((5 r^2 sin 72°)/2)−((a^2 sin 72°)/2) ((shaded)/(total))=((5r^2 −a^2 )/(5R^2 +a^2 ))=((5((r/a))^2 −1)/(5((R/a))^2 +1)) =((5((1/(2 sin 36°)))^2 −1)/(5(((sin 54°)/(sin 36°)))^2 +1))=(1/4)=25% ⇒answer b)

$${a}={side}\:{length}\:{of}\:{small}\:{pentagon} \\ $$$${b}={side}\:{length}\:{of}\:{big}\:{pentagon} \\ $$$${b}=\mathrm{2}{a}\:\mathrm{sin}\:\mathrm{54}° \\ $$$${a}=\mathrm{2}{r}\:\mathrm{sin}\:\mathrm{36}°\:\Rightarrow{r}=\frac{{a}}{\mathrm{2}\:\mathrm{sin}\:\mathrm{36}°} \\ $$$${b}=\mathrm{2}{R}\:\mathrm{sin}\:\mathrm{36}°\:\Rightarrow{R}=\frac{{b}}{\mathrm{2}\:\mathrm{sin}\:\mathrm{36}°}=\frac{{a}\:\mathrm{sin}\:\mathrm{54}°}{\mathrm{sin}\:\mathrm{36}°} \\ $$$${total}\:{area}\:=\frac{\mathrm{5}\:{R}^{\mathrm{2}} \:\mathrm{sin}\:\mathrm{72}°}{\mathrm{2}}+\frac{{a}^{\mathrm{2}} \mathrm{sin}\:\mathrm{72}°}{\mathrm{2}} \\ $$$${shaded}\:{area}=\frac{\mathrm{5}\:{r}^{\mathrm{2}} \:\mathrm{sin}\:\mathrm{72}°}{\mathrm{2}}−\frac{{a}^{\mathrm{2}} \mathrm{sin}\:\mathrm{72}°}{\mathrm{2}} \\ $$$$\frac{{shaded}}{{total}}=\frac{\mathrm{5}{r}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{5}{R}^{\mathrm{2}} +{a}^{\mathrm{2}} }=\frac{\mathrm{5}\left(\frac{{r}}{{a}}\right)^{\mathrm{2}} −\mathrm{1}}{\mathrm{5}\left(\frac{{R}}{{a}}\right)^{\mathrm{2}} +\mathrm{1}} \\ $$$$\:\:=\frac{\mathrm{5}\left(\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{sin}\:\mathrm{36}°}\right)^{\mathrm{2}} −\mathrm{1}}{\mathrm{5}\left(\frac{\mathrm{sin}\:\mathrm{54}°}{\mathrm{sin}\:\mathrm{36}°}\right)^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{25\%} \\ $$$$\left.\Rightarrow{answer}\:{b}\right) \\ $$

Commented by Mingma last updated on 06/Aug/23

Perfect ��

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