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

$${especially}\:{MjS}\:{Sir}\:.. \\ $$

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

$${yes}. \\ $$

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

$${equal}\:{sided}\:{only},\:{given}\:{in}\:{question}. \\ $$

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

$${Very}\:{well}\:{Sir},\:{whats}\:{this}\:{app}\:{you} \\ $$$${graph}\:{on},\:{thank}\:{you}. \\ $$

Commented by mr W last updated on 29/Dec/18

Commented by mr W last updated on 29/Dec/18

$${it}'{s}\:{a}\:{free}\:{app}\:{which}\:{you}\:{can}\:{download} \\ $$$${from}\:{google}\:{play}.\:{a}\:{very}\:{very}\:{good}\:{app}! \\ $$

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

$${Lekh}\:{Diagram} \\ $$

Commented by afachri last updated on 31/Dec/18

$$\mathrm{thNks}\:\mathrm{Mr}\:\mathrm{Ajfour} \\ $$

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