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 (√((R−a)^2 −a^2 ))+(√((R−b)^2 −b^2 ))=a+b (√(R(R−2a)))=a+b−(√(R(R−2b))) R(R−2a)=(a+b)^2 +R(R−2b)−2(a+b)(√(R(R−2b))) (a+b)^2 −2(b−a)R=2(a+b)(√(R(R−2b))) (a+b)^2 −4(b−a)(a+b)^2 R+4(b−a)^2 R^2 =4(a+b)^2 R^2 −8b(a+b)^2 R 16abR^2 −4(a+b)^3 R−(a+b)^4 =0 R=(((a+b)^2 [a+b+(√((a+b)^2 +4ab))])/(8ab)) AB=2R ⇒AB=(((a+b)^2 [a+b+(√((a+b)^2 +4ab))])/(4ab)) with a=4, b=6 AB=((100(10+(√(100+4×24))))/(4×24))=25](https://www.tinkutara.com/question/Q191724.png)
$${R}={radius}\:{of}\:{semi}−{circle} \\ $$$$\sqrt{\left({R}−{a}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} }+\sqrt{\left({R}−{b}\right)^{\mathrm{2}} −{b}^{\mathrm{2}} }={a}+{b} \\ $$$$\sqrt{{R}\left({R}−\mathrm{2}{a}\right)}={a}+{b}−\sqrt{{R}\left({R}−\mathrm{2}{b}\right)} \\ $$$${R}\left({R}−\mathrm{2}{a}\right)=\left({a}+{b}\right)^{\mathrm{2}} +{R}\left({R}−\mathrm{2}{b}\right)−\mathrm{2}\left({a}+{b}\right)\sqrt{{R}\left({R}−\mathrm{2}{b}\right)} \\ $$$$\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{2}\left({b}−{a}\right){R}=\mathrm{2}\left({a}+{b}\right)\sqrt{{R}\left({R}−\mathrm{2}{b}\right)} \\ $$$$\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{4}\left({b}−{a}\right)\left({a}+{b}\right)^{\mathrm{2}} {R}+\mathrm{4}\left({b}−{a}\right)^{\mathrm{2}} {R}^{\mathrm{2}} =\mathrm{4}\left({a}+{b}\right)^{\mathrm{2}} {R}^{\mathrm{2}} −\mathrm{8}{b}\left({a}+{b}\right)^{\mathrm{2}} {R} \\ $$$$\mathrm{16}{abR}^{\mathrm{2}} −\mathrm{4}\left({a}+{b}\right)^{\mathrm{3}} {R}−\left({a}+{b}\right)^{\mathrm{4}} =\mathrm{0} \\ $$$${R}=\frac{\left({a}+{b}\right)^{\mathrm{2}} \left[{a}+{b}+\sqrt{\left({a}+{b}\right)^{\mathrm{2}} +\mathrm{4}{ab}}\right]}{\mathrm{8}{ab}} \\ $$$${AB}=\mathrm{2}{R} \\ $$$$\Rightarrow{AB}=\frac{\left({a}+{b}\right)^{\mathrm{2}} \left[{a}+{b}+\sqrt{\left({a}+{b}\right)^{\mathrm{2}} +\mathrm{4}{ab}}\right]}{\mathrm{4}{ab}} \\ $$$${with}\:{a}=\mathrm{4},\:{b}=\mathrm{6} \\ $$$${AB}=\frac{\mathrm{100}\left(\mathrm{10}+\sqrt{\mathrm{100}+\mathrm{4}×\mathrm{24}}\right)}{\mathrm{4}×\mathrm{24}}=\mathrm{25} \\ $$
Commented by Shlock last updated on 29/Apr/23
Perfect ��