Question Number 224118 by fantastic last updated on 21/Aug/25
Answered by mr W last updated on 22/Aug/25
$${R}=\mathrm{12} \\ $$$$\mathrm{2}{b}={R}\:\Rightarrow{b}=\frac{{R}}{\mathrm{2}} \\ $$$$\left({a}+{b}\right)^{\mathrm{2}} ={b}^{\mathrm{2}} +\left({R}−{a}\right)^{\mathrm{2}} \\ $$$$\mathrm{2}\left({R}+{b}\right){a}={R}^{\mathrm{2}} \\ $$$$\Rightarrow{a}=\frac{{R}^{\mathrm{2}} }{\mathrm{2}\left({R}+{b}\right)}=\frac{{R}}{\mathrm{3}} \\ $$$$\left({R}−{c}\right)^{\mathrm{2}} −{c}^{\mathrm{2}} =\left({b}+{c}\right)^{\mathrm{2}} −\left({b}−{c}\right)^{\mathrm{2}} \\ $$$${R}^{\mathrm{2}} =\mathrm{2}\left({R}+\mathrm{2}{b}\right){c} \\ $$$$\Rightarrow{c}=\frac{{R}^{\mathrm{2}} }{\mathrm{2}\left({R}+\mathrm{2}{b}\right)}=\frac{{R}}{\mathrm{4}} \\ $$$${a}+{b}+{c}=\frac{{R}}{\mathrm{3}}+\frac{{R}}{\mathrm{2}}+\frac{{R}}{\mathrm{4}}=\frac{\mathrm{13}{R}}{\mathrm{12}}=\mathrm{13}\:\checkmark \\ $$
Commented by fantastic last updated on 22/Aug/25
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