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Question Number 188378 by Mingma last updated on 28/Feb/23

Answered by HeferH last updated on 09/Mar/23

$$letxbethesideoftheredtriangleandybethe$$$$sideoftheblueone$$$$\sin \left(18°\right)=\frac{y}{x}$$$$\frac{A_r}{A_b}={\left(\frac{1}{\sin \left(18°\right)}\right)}^2={\left(\frac{x}{y}\right)}^2={\left(\frac{4}{\sqrt{5}-1}\right)}^2=\frac{8}{3-\sqrt{5}}$$$$=\frac{8(3+\sqrt{5})}{9-5}=2(3+\sqrt{5})=6+2\sqrt{5}$$

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