Question Number 191715 by Shlock last updated on 29/Apr/23
Answered by mr W last updated on 29/Apr/23
Commented by Shlock last updated on 29/Apr/23
Sir,canyon use the similarity of the right triangles?
Can you give more explanation in details?
Commented by mr W last updated on 29/Apr/23
$${R}={radius}\:{of}\:{semi}−{circle} \\ $$$$\sqrt{\left({R}−{a}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} }+\sqrt{\left({R}−{b}\right)^{\mathrm{2}} −{b}^{\mathrm{2}} }={a}+{b} \\ $$$$\sqrt{{R}\left({R}−\mathrm{2}{a}\right)}={a}+{b}−\sqrt{{R}\left({R}−\mathrm{2}{b}\right)} \\ $$$${R}\left({R}−\mathrm{2}{a}\right)=\left({a}+{b}\right)^{\mathrm{2}} +{R}\left({R}−\mathrm{2}{b}\right)−\mathrm{2}\left({a}+{b}\right)\sqrt{{R}\left({R}−\mathrm{2}{b}\right)} \\ $$$$\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{2}\left({b}−{a}\right){R}=\mathrm{2}\left({a}+{b}\right)\sqrt{{R}\left({R}−\mathrm{2}{b}\right)} \\ $$$$\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{4}\left({b}−{a}\right)\left({a}+{b}\right)^{\mathrm{2}} {R}+\mathrm{4}\left({b}−{a}\right)^{\mathrm{2}} {R}^{\mathrm{2}} =\mathrm{4}\left({a}+{b}\right)^{\mathrm{2}} {R}^{\mathrm{2}} −\mathrm{8}{b}\left({a}+{b}\right)^{\mathrm{2}} {R} \\ $$$$\mathrm{16}{abR}^{\mathrm{2}} −\mathrm{4}\left({a}+{b}\right)^{\mathrm{3}} {R}−\left({a}+{b}\right)^{\mathrm{4}} =\mathrm{0} \\ $$$${R}=\frac{\left({a}+{b}\right)^{\mathrm{2}} \left[{a}+{b}+\sqrt{\left({a}+{b}\right)^{\mathrm{2}} +\mathrm{4}{ab}}\right]}{\mathrm{8}{ab}} \\ $$$${AB}=\mathrm{2}{R} \\ $$$$\Rightarrow{AB}=\frac{\left({a}+{b}\right)^{\mathrm{2}} \left[{a}+{b}+\sqrt{\left({a}+{b}\right)^{\mathrm{2}} +\mathrm{4}{ab}}\right]}{\mathrm{4}{ab}} \\ $$$${with}\:{a}=\mathrm{4},\:{b}=\mathrm{6} \\ $$$${AB}=\frac{\mathrm{100}\left(\mathrm{10}+\sqrt{\mathrm{100}+\mathrm{4}×\mathrm{24}}\right)}{\mathrm{4}×\mathrm{24}}=\mathrm{25} \\ $$
Commented by Shlock last updated on 29/Apr/23
Perfect