Question-65366

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

$$\mathrm{the}\:\mathrm{length}\:\mathrm{is} \\ $$$$=\:\mathrm{3}\:+\:\mathrm{2}×\mathrm{2}\:+\mathrm{2}×\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{3}}\:+\:\mathrm{2}×\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{2}} }\:+\:\mathrm{2}×\frac{\mathrm{2}^{\mathrm{4}} }{\mathrm{3}^{\mathrm{3}} }\:+\:……. \\ $$$$=\mathrm{3}+\mathrm{4}\left[\mathrm{1}+\:\left(\frac{\mathrm{2}}{\mathrm{3}}\right)+\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{3}} +……..\right] \\ $$$$=\:\mathrm{3}\:+\:\mathrm{4}\left[\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}}\right]\:=\:\mathrm{3}+\:\mathrm{4}×\mathrm{3}\:=\mathrm{15} \\ $$

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 ≡ horizontal side b_n ≡ vertical side = (2/3)a_n a_(n+1) = b_n l_n = a_n + b_n l_(n+1) =(2/3)l_n triangle #1: l_1 = 3+2 = 5 triangle #2: l_2 = 2+2(2/3)=((10)/3)=(2/3)×5=(2/3)l_1 triangle #3: l_3 = (4/3)+(4/3)×(2/3)=((20)/9)=(2/3)×((10)/3)=(2/3)l_2 this is a geometric progression

$${a}_{{n}} \equiv\:{horizontal}\:{side} \\ $$$${b}_{{n}} \equiv\:{vertical}\:{side}\:=\:\frac{\mathrm{2}}{\mathrm{3}}{a}_{{n}} \\ $$$${a}_{{n}+\mathrm{1}} \:=\:{b}_{{n}} \:\:\:\:\:\:{l}_{{n}} \:=\:{a}_{{n}} +\:{b}_{{n}} \:\:\:\:\:\:{l}_{{n}+\mathrm{1}} =\frac{\mathrm{2}}{\mathrm{3}}{l}_{{n}} \\ $$$$ \\ $$$${triangle}\:#\mathrm{1}:\:{l}_{\mathrm{1}} \:=\:\:\mathrm{3}+\mathrm{2}\:=\:\mathrm{5} \\ $$$${triangle}\:#\mathrm{2}:\:{l}_{\mathrm{2}} \:=\:\mathrm{2}+\mathrm{2}\frac{\mathrm{2}}{\mathrm{3}}=\frac{\mathrm{10}}{\mathrm{3}}=\frac{\mathrm{2}}{\mathrm{3}}×\mathrm{5}=\frac{\mathrm{2}}{\mathrm{3}}{l}_{\mathrm{1}} \\ $$$${triangle}\:#\mathrm{3}:\:{l}_{\mathrm{3}} \:=\:\frac{\mathrm{4}}{\mathrm{3}}+\frac{\mathrm{4}}{\mathrm{3}}×\frac{\mathrm{2}}{\mathrm{3}}=\frac{\mathrm{20}}{\mathrm{9}}=\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{10}}{\mathrm{3}}=\frac{\mathrm{2}}{\mathrm{3}}{l}_{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{a}\:\boldsymbol{\mathrm{geometric}}\:\boldsymbol{\mathrm{progression}} \\ $$

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+3 (a/b)=((b+3)/b)=(3/2) ⇒1+(3/b)=(3/2) ⇒b=6 ⇒a=6+3=9 length of red path=a+b=9+6=15

$${a}={b}+\mathrm{3} \\ $$$$\frac{{a}}{{b}}=\frac{{b}+\mathrm{3}}{{b}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{1}+\frac{\mathrm{3}}{{b}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow{b}=\mathrm{6} \\ $$$$\Rightarrow{a}=\mathrm{6}+\mathrm{3}=\mathrm{9} \\ $$$${length}\:{of}\:{red}\:{path}={a}+{b}=\mathrm{9}+\mathrm{6}=\mathrm{15} \\ $$

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 (a+b). the steps may be irregular. for example the red path, the blue path and the green path all have the same length.

$${for}\:{a}\:{zig}−{zag}\:{path}\:{the}\:{path}\:{length}\:{is} \\ $$$${the}\:{sum}\:{of}\:{horizontal}\:{and}\:{vertical} \\ $$$${projection}\:\left({a}+{b}\right).\:{the}\:{steps}\:{may}\:{be} \\ $$$${irregular}.\:{for}\:{example}\:{the}\:{red}\:{path}, \\ $$$${the}\:{blue}\:{path}\:{and}\:{the}\:{green}\:{path}\:{all} \\ $$$${have}\:{the}\:{same}\:{length}. \\ $$

Commented by mr W last updated on 29/Jul/19