Question Number 65366 by Tawa1 last updated on 29/Jul/19
Commented by Prithwish sen last updated on 29/Jul/19
![the length is = 3 + 2×2 +2×(2^2 /3) + 2×(2^3 /3^2 ) + 2×(2^4 /3^3 ) + ....... =3+4[1+ ((2/3))+((2/3))^2 +((2/3))^3 +........] = 3 + 4[(1/(1−(2/3)))] = 3+ 4×3 =15](Q65412.png)
$$\mathrm{the}\:\mathrm{length}\:\mathrm{is} \\ $$$$=\:\mathrm{3}\:+\:\mathrm{2}×\mathrm{2}\:+\mathrm{2}×\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{3}}\:+\:\mathrm{2}×\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{2}} }\:+\:\mathrm{2}×\frac{\mathrm{2}^{\mathrm{4}} }{\mathrm{3}^{\mathrm{3}} }\:+\:....... \\ $$$$=\mathrm{3}+\mathrm{4}\left[\mathrm{1}+\:\left(\frac{\mathrm{2}}{\mathrm{3}}\right)+\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{3}} +........\right] \\ $$$$=\:\mathrm{3}\:+\:\mathrm{4}\left[\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}}\right]\:=\:\mathrm{3}+\:\mathrm{4}×\mathrm{3}\:=\mathrm{15} \\ $$
Commented by Tawa1 last updated on 29/Jul/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by JDamian last updated on 29/Jul/19
$${L}=\frac{\mathrm{2}+\mathrm{3}}{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}}=\frac{\mathrm{5}}{\frac{\mathrm{1}}{\mathrm{3}}}=\mathrm{15}\:{feet} \\ $$
Commented by KanhAshish last updated on 29/Jul/19
$$\mathrm{how}?? \\ $$
Commented by JDamian last updated on 29/Jul/19
$${a}_{{n}} \equiv\:{horizontal}\:{side} \\ $$$${b}_{{n}} \equiv\:{vertical}\:{side}\:=\:\frac{\mathrm{2}}{\mathrm{3}}{a}_{{n}} \\ $$$${a}_{{n}+\mathrm{1}} \:=\:{b}_{{n}} \:\:\:\:\:\:{l}_{{n}} \:=\:{a}_{{n}} +\:{b}_{{n}} \:\:\:\:\:\:{l}_{{n}+\mathrm{1}} =\frac{\mathrm{2}}{\mathrm{3}}{l}_{{n}} \\ $$$$ \\ $$$${triangle}\:#\mathrm{1}:\:{l}_{\mathrm{1}} \:=\:\:\mathrm{3}+\mathrm{2}\:=\:\mathrm{5} \\ $$$${triangle}\:#\mathrm{2}:\:{l}_{\mathrm{2}} \:=\:\mathrm{2}+\mathrm{2}\frac{\mathrm{2}}{\mathrm{3}}=\frac{\mathrm{10}}{\mathrm{3}}=\frac{\mathrm{2}}{\mathrm{3}}×\mathrm{5}=\frac{\mathrm{2}}{\mathrm{3}}{l}_{\mathrm{1}} \\ $$$${triangle}\:#\mathrm{3}:\:{l}_{\mathrm{3}} \:=\:\frac{\mathrm{4}}{\mathrm{3}}+\frac{\mathrm{4}}{\mathrm{3}}×\frac{\mathrm{2}}{\mathrm{3}}=\frac{\mathrm{20}}{\mathrm{9}}=\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{10}}{\mathrm{3}}=\frac{\mathrm{2}}{\mathrm{3}}{l}_{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{a}\:\boldsymbol{\mathrm{geometric}}\:\boldsymbol{\mathrm{progression}} \\ $$
Commented by KanhAshish last updated on 29/Jul/19
$$\mathrm{amazing}\:\mathrm{explanation}..\:\mathrm{god}\:\mathrm{bless}\:\mathrm{you}. \\ $$
Commented by Tawa1 last updated on 29/Jul/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by mr W last updated on 29/Jul/19
Commented by mr W last updated on 29/Jul/19
$${a}={b}+\mathrm{3} \\ $$$$\frac{{a}}{{b}}=\frac{{b}+\mathrm{3}}{{b}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{1}+\frac{\mathrm{3}}{{b}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow{b}=\mathrm{6} \\ $$$$\Rightarrow{a}=\mathrm{6}+\mathrm{3}=\mathrm{9} \\ $$$${length}\:{of}\:{red}\:{path}={a}+{b}=\mathrm{9}+\mathrm{6}=\mathrm{15} \\ $$
Commented by mr W last updated on 29/Jul/19
$${for}\:{a}\:{zig}−{zag}\:{path}\:{the}\:{path}\:{length}\:{is} \\ $$$${the}\:{sum}\:{of}\:{horizontal}\:{and}\:{vertical} \\ $$$${projection}\:\left({a}+{b}\right).\:{the}\:{steps}\:{may}\:{be} \\ $$$${irregular}.\:{for}\:{example}\:{the}\:{red}\:{path}, \\ $$$${the}\:{blue}\:{path}\:{and}\:{the}\:{green}\:{path}\:{all} \\ $$$${have}\:{the}\:{same}\:{length}. \\ $$
Commented by mr W last updated on 29/Jul/19
Commented by Tawa1 last updated on 29/Jul/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{appreciate} \\ $$