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Question-185768




Question Number 185768 by Mingma last updated on 27/Jan/23
Answered by JDamian last updated on 27/Jan/23
l=(2/3)L  outer triangle area         A=kL^2   inner triangle area           a=kl^2 =k((2/3)L)^2 =k(4/9)L^2     (A/a)=((kL^2 )/(k(4/9)L^2 ))=(9/4)  ⇒  A=(9/4)a    trapezoid area  s=((A−a)/3)=(((9/4)a−a)/3)=(5/(12))a    (s/a)=(5/(12))
$${l}=\frac{\mathrm{2}}{\mathrm{3}}{L} \\ $$$$\mathrm{outer}\:\mathrm{triangle}\:\mathrm{area} \\ $$$$\:\:\:\:\:\:\:{A}={kL}^{\mathrm{2}} \\ $$$$\mathrm{inner}\:\mathrm{triangle}\:\mathrm{area} \\ $$$$\:\:\:\:\:\:\:\:\:{a}={kl}^{\mathrm{2}} ={k}\left(\frac{\mathrm{2}}{\mathrm{3}}{L}\right)^{\mathrm{2}} ={k}\frac{\mathrm{4}}{\mathrm{9}}{L}^{\mathrm{2}} \\ $$$$ \\ $$$$\frac{{A}}{{a}}=\frac{{kL}^{\mathrm{2}} }{{k}\frac{\mathrm{4}}{\mathrm{9}}{L}^{\mathrm{2}} }=\frac{\mathrm{9}}{\mathrm{4}}\:\:\Rightarrow\:\:{A}=\frac{\mathrm{9}}{\mathrm{4}}{a} \\ $$$$ \\ $$$$\mathrm{trapezoid}\:\mathrm{area} \\ $$$${s}=\frac{{A}−{a}}{\mathrm{3}}=\frac{\frac{\mathrm{9}}{\mathrm{4}}{a}−{a}}{\mathrm{3}}=\frac{\mathrm{5}}{\mathrm{12}}{a} \\ $$$$ \\ $$$$\frac{{s}}{{a}}=\frac{\mathrm{5}}{\mathrm{12}} \\ $$

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