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Question-51658




Question Number 51658 by ajfour last updated on 29/Dec/18
Commented by ajfour last updated on 29/Dec/18
Find in terms of a and b the side  length s of the equal sided pentagon.
$${Find}\:{in}\:{terms}\:{of}\:{a}\:{and}\:{b}\:{the}\:{side} \\ $$$${length}\:\boldsymbol{{s}}\:{of}\:{the}\:{equal}\:{sided}\:{pentagon}. \\ $$
Commented by ajfour last updated on 29/Dec/18
especially MjS Sir ..
$${especially}\:{MjS}\:{Sir}\:.. \\ $$
Commented by afachri last updated on 29/Dec/18
all pentagon sides have the same S lenght  sir ???
$$\mathrm{all}\:\mathrm{pentagon}\:\mathrm{sides}\:\mathrm{have}\:\mathrm{the}\:\mathrm{same}\:\mathrm{S}\:\mathrm{lenght} \\ $$$$\mathrm{sir}\:??? \\ $$
Commented by ajfour last updated on 29/Dec/18
yes.
$${yes}. \\ $$
Commented by afachri last updated on 29/Dec/18
i have an answer Sir. Wait, i′ll post it  and please correct me if i am mistaken.
$$\mathrm{i}\:\mathrm{have}\:\mathrm{an}\:\mathrm{answer}\:\mathrm{Sir}.\:\mathrm{Wait},\:\mathrm{i}'\mathrm{ll}\:\mathrm{post}\:\mathrm{it} \\ $$$$\mathrm{and}\:\mathrm{please}\:\mathrm{correct}\:\mathrm{me}\:\mathrm{if}\:\mathrm{i}\:\mathrm{am}\:\mathrm{mistaken}. \\ $$
Commented by Kunal12588 last updated on 29/Dec/18
sir is it a equilateral pentagon with non equal  angles? or a regular pentagon(it will be easier) ?
$${sir}\:{is}\:{it}\:{a}\:{equilateral}\:{pentagon}\:{with}\:{non}\:{equal} \\ $$$${angles}?\:{or}\:{a}\:{regular}\:{pentagon}\left({it}\:{will}\:{be}\:{easier}\right)\:? \\ $$
Commented by ajfour last updated on 29/Dec/18
equal sided only, given in question.
$${equal}\:{sided}\:{only},\:{given}\:{in}\:{question}. \\ $$
Answered by afachri last updated on 29/Dec/18
Commented by mr W last updated on 29/Dec/18
it′s not correct sir.  you assume a=2x+s and b=x+y. but  this is not true. the triangle on point  C is not the same as the triangle on  point B. otherwise DC=a=2y!
$${it}'{s}\:{not}\:{correct}\:{sir}. \\ $$$${you}\:{assume}\:{a}=\mathrm{2}{x}+{s}\:{and}\:{b}={x}+{y}.\:{but} \\ $$$${this}\:{is}\:{not}\:{true}.\:{the}\:{triangle}\:{on}\:{point} \\ $$$${C}\:{is}\:{not}\:{the}\:{same}\:{as}\:{the}\:{triangle}\:{on} \\ $$$${point}\:{B}.\:{otherwise}\:{DC}={a}=\mathrm{2}{y}! \\ $$
Commented by afachri last updated on 29/Dec/18
thanks for your correction Mr W.   so that i thought DC = a = 2y.   i′ll fix it Mr W. really appreciate your  correction Sir.
$$\mathrm{thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{correction}\:\mathrm{Mr}\:\mathrm{W}.\: \\ $$$$\mathrm{so}\:\mathrm{that}\:\mathrm{i}\:\mathrm{thought}\:{DC}\:=\:{a}\:=\:\mathrm{2}{y}.\: \\ $$$$\mathrm{i}'\mathrm{ll}\:\mathrm{fix}\:\mathrm{it}\:\mathrm{Mr}\:\mathrm{W}.\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{your} \\ $$$$\mathrm{correction}\:\mathrm{Sir}. \\ $$
Answered by afachri last updated on 29/Dec/18
side a        2x + s  =  a                                      x  =  ((a−s)/2)  side b        x + y     =  b                                      y  =  b − x                                      y  =   b − ((a−s)/2)  based on phytagoras theorm :   s^2 =  x^2 + y^2                                         s^2                      =  (((a−s)/2))^2 + (b − ((a−s)/2))^2                                          s^2                      =  (((a−s)^2 )/4) + b^2 − b(a−s) + (((a−s)^2 )/4)                                        s^2                     =  b^2 − ab + bs + (((a−s)^2 )/2)  then multiple by 2                                      2s^2                    =  2b^2 − 2ab + 2bs + (a^2 − 2as + s^2 )                                       2s^2                    =  s^2 + 2(b−a)s + (a^2 + 2b^2 − 2ab)                          s^2 − 2(b − a)s  =  a^2 + 2b^2 − 2ab  s^2 − 2(b−a)s + (b−a)^2   =  a^2 + 2b^2 − 2ab + (b−a)^2                             [ s − (b−a) ]^2   =  2a^2 − 4ab + 3b^2                             [ s − (b−a) ]^2   = 2(a−b)^2 + b^2                                     s − (b−a)  =  (√(2(a−b)^2 + b^2  ))                                    s                          =   (b−a) + (√(2(a−b)^2 + b^2   ))
$$\boldsymbol{\mathrm{side}}\:\boldsymbol{{a}}\:\:\:\:\:\:\:\:\mathrm{2}\boldsymbol{{x}}\:+\:\boldsymbol{{s}}\:\:=\:\:\boldsymbol{{a}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{x}}\:\:=\:\:\frac{\boldsymbol{{a}}−\boldsymbol{{s}}}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{side}}\:\boldsymbol{{b}}\:\:\:\:\:\:\:\:\boldsymbol{{x}}\:+\:\boldsymbol{{y}}\:\:\:\:\:=\:\:\boldsymbol{{b}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{y}}\:\:=\:\:\boldsymbol{{b}}\:−\:\boldsymbol{{x}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{y}}\:\:=\:\:\:\boldsymbol{{b}}\:−\:\frac{\boldsymbol{{a}}−\boldsymbol{{s}}}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{based}}\:\boldsymbol{\mathrm{on}}\:\boldsymbol{\mathrm{phytagoras}}\:\boldsymbol{\mathrm{theorm}}\::\:\:\:\boldsymbol{{s}}^{\mathrm{2}} =\:\:\boldsymbol{{x}}^{\mathrm{2}} +\:\boldsymbol{{y}}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{s}}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\left(\frac{\boldsymbol{{a}}−\boldsymbol{{s}}}{\mathrm{2}}\right)^{\mathrm{2}} +\:\left(\boldsymbol{{b}}\:−\:\frac{\boldsymbol{{a}}−\boldsymbol{{s}}}{\mathrm{2}}\right)^{\mathrm{2}} \: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{s}}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\frac{\left(\boldsymbol{{a}}−\boldsymbol{{s}}\right)^{\mathrm{2}} }{\mathrm{4}}\:+\:\boldsymbol{{b}}^{\mathrm{2}} −\:\boldsymbol{{b}}\left(\boldsymbol{{a}}−\boldsymbol{{s}}\right)\:+\:\frac{\left(\boldsymbol{{a}}−\boldsymbol{{s}}\right)^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{s}}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\boldsymbol{{b}}^{\mathrm{2}} −\:\boldsymbol{{ab}}\:+\:\boldsymbol{{bs}}\:+\:\frac{\left(\boldsymbol{{a}}−\boldsymbol{{s}}\right)^{\mathrm{2}} }{\mathrm{2}}\:\:\boldsymbol{\mathrm{then}}\:\boldsymbol{\mathrm{multiple}}\:\boldsymbol{\mathrm{by}}\:\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\boldsymbol{{s}}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\mathrm{2}\boldsymbol{{b}}^{\mathrm{2}} −\:\mathrm{2}\boldsymbol{{ab}}\:+\:\mathrm{2}\boldsymbol{{bs}}\:+\:\left(\boldsymbol{{a}}^{\mathrm{2}} −\:\mathrm{2}\boldsymbol{{as}}\:+\:\boldsymbol{{s}}^{\mathrm{2}} \right)\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\boldsymbol{{s}}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\boldsymbol{{s}}^{\mathrm{2}} +\:\mathrm{2}\left(\boldsymbol{{b}}−\boldsymbol{{a}}\right)\boldsymbol{{s}}\:+\:\left(\boldsymbol{{a}}^{\mathrm{2}} +\:\mathrm{2}\boldsymbol{{b}}^{\mathrm{2}} −\:\mathrm{2}\boldsymbol{{ab}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{s}}^{\mathrm{2}} −\:\mathrm{2}\left(\boldsymbol{{b}}\:−\:\boldsymbol{{a}}\right)\boldsymbol{{s}}\:\:=\:\:\boldsymbol{{a}}^{\mathrm{2}} +\:\mathrm{2}\boldsymbol{{b}}^{\mathrm{2}} −\:\mathrm{2}\boldsymbol{{ab}} \\ $$$$\boldsymbol{{s}}^{\mathrm{2}} −\:\mathrm{2}\left(\boldsymbol{{b}}−\boldsymbol{{a}}\right)\boldsymbol{{s}}\:+\:\left(\boldsymbol{{b}}−\boldsymbol{{a}}\right)^{\mathrm{2}} \:\:=\:\:\boldsymbol{{a}}^{\mathrm{2}} +\:\mathrm{2}\boldsymbol{{b}}^{\mathrm{2}} −\:\mathrm{2}\boldsymbol{{ab}}\:+\:\left(\boldsymbol{{b}}−\boldsymbol{{a}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\:\boldsymbol{{s}}\:−\:\left(\boldsymbol{{b}}−\boldsymbol{{a}}\right)\:\right]^{\mathrm{2}} \:\:=\:\:\mathrm{2}\boldsymbol{{a}}^{\mathrm{2}} −\:\mathrm{4}\boldsymbol{{ab}}\:+\:\mathrm{3}\boldsymbol{{b}}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\:\boldsymbol{{s}}\:−\:\left(\boldsymbol{{b}}−\boldsymbol{{a}}\right)\:\right]^{\mathrm{2}} \:\:=\:\mathrm{2}\left(\boldsymbol{{a}}−\boldsymbol{{b}}\right)^{\mathrm{2}} +\:\boldsymbol{{b}}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{s}}\:−\:\left(\boldsymbol{{b}}−\boldsymbol{{a}}\right)\:\:=\:\:\sqrt{\mathrm{2}\left(\boldsymbol{{a}}−\boldsymbol{{b}}\right)^{\mathrm{2}} +\:\boldsymbol{{b}}^{\mathrm{2}} \:} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{s}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\:\left(\boldsymbol{{b}}−\boldsymbol{{a}}\right)\:+\:\sqrt{\mathrm{2}\left(\boldsymbol{{a}}−\boldsymbol{{b}}\right)^{\mathrm{2}} +\:\boldsymbol{{b}}^{\mathrm{2}} \:\:} \\ $$$$ \\ $$
Answered by mr W last updated on 29/Dec/18
(√(s^2 −(a^2 /4)))+(√(s^2 −(((a−s)/2))^2 ))=b  with μ=(b/a) and λ=(s/a)  ⇒(√(λ^2 −(1/4)))+(√(λ^2 −(((1−λ)/2))^2 ))=μ  ⇒λ=f(μ) (only numerically, I think)    examples:  b=0.5a⇒s≈0.5038a  b=a⇒s≈0.6351a  b=2a⇒s≈1.0627a
$$\sqrt{{s}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}+\sqrt{{s}^{\mathrm{2}} −\left(\frac{{a}−{s}}{\mathrm{2}}\right)^{\mathrm{2}} }={b} \\ $$$${with}\:\mu=\frac{{b}}{{a}}\:{and}\:\lambda=\frac{{s}}{{a}} \\ $$$$\Rightarrow\sqrt{\lambda^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}}+\sqrt{\lambda^{\mathrm{2}} −\left(\frac{\mathrm{1}−\lambda}{\mathrm{2}}\right)^{\mathrm{2}} }=\mu \\ $$$$\Rightarrow\lambda={f}\left(\mu\right)\:\left({only}\:{numerically},\:{I}\:{think}\right) \\ $$$$ \\ $$$${examples}: \\ $$$${b}=\mathrm{0}.\mathrm{5}{a}\Rightarrow{s}\approx\mathrm{0}.\mathrm{5038}{a} \\ $$$${b}={a}\Rightarrow{s}\approx\mathrm{0}.\mathrm{6351}{a} \\ $$$${b}=\mathrm{2}{a}\Rightarrow{s}\approx\mathrm{1}.\mathrm{0627}{a} \\ $$
Commented by mr W last updated on 29/Dec/18
Commented by ajfour last updated on 29/Dec/18
Very well Sir, whats this app you  graph on, thank you.
$${Very}\:{well}\:{Sir},\:{whats}\:{this}\:{app}\:{you} \\ $$$${graph}\:{on},\:{thank}\:{you}. \\ $$
Commented by mr W last updated on 29/Dec/18
Commented by mr W last updated on 29/Dec/18
it′s a free app which you can download  from google play. a very very good app!
$${it}'{s}\:{a}\:{free}\:{app}\:{which}\:{you}\:{can}\:{download} \\ $$$${from}\:{google}\:{play}.\:{a}\:{very}\:{very}\:{good}\:{app}! \\ $$
Commented by afachri last updated on 30/Dec/18
what about an app used for draw in geometry ??  any suggestion, Sir ?
$$\mathrm{what}\:\mathrm{about}\:\mathrm{an}\:\mathrm{app}\:\mathrm{used}\:\mathrm{for}\:\mathrm{draw}\:\mathrm{in}\:\mathrm{geometry}\:?? \\ $$$$\mathrm{any}\:\mathrm{suggestion},\:\mathrm{Sir}\:? \\ $$
Commented by ajfour last updated on 30/Dec/18
Lekh Diagram
$${Lekh}\:{Diagram} \\ $$
Commented by afachri last updated on 31/Dec/18
thNks Mr Ajfour
$$\mathrm{thNks}\:\mathrm{Mr}\:\mathrm{Ajfour} \\ $$

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