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Question Number 49536 by behi83417@gmail.com last updated on 07/Dec/18

Commented by behi83417@gmail.com last updated on 07/Dec/18

△ and△ are equilateral triangles with parallel sides and PM,is the same for all sides.if:area(△)=32 and area(△)=6, then:valve of :PM=?

$$\bigtriangleup\:{and}\bigtriangleup\:{are}\:{equilateral}\:{triangles}\:{with} \\ $$$${parallel}\:{sides}\:{and}\:\boldsymbol{\mathrm{PM}},{is}\:{the}\:{same}\:{for} \\ $$$${all}\:{sides}.{if}:{area}\left(\bigtriangleup\right)=\mathrm{32}\:{and}\:{area}\left(\bigtriangleup\right)=\mathrm{6}, \\ $$$${then}:{valve}\:{of}\::\boldsymbol{\mathrm{PM}}=? \\ $$

Commented by Kunal12588 last updated on 07/Dec/18

is answer 1.40704302?

Answered by mr W last updated on 07/Dec/18

A=(((√3)a^2 )/4) ⇒a=(√((4A)/(√3)))=((2(√(3(√3)A)))/3) a_(red) =((2(√(3(√3)×32)))/3)=((8(√(6(√3))))/3) a_(blue) =((2(√(3(√3)×6)))/3)=2(√(2(√3))) let δ=PM a_(red) −a_(blue) =2(√3)δ ((8(√(6(√3))))/3)−2(√(2(√3)))=2(√3)δ ⇒δ=((4(√(2(√3)))−(√(6(√3))))/3)≈1.407

$${A}=\frac{\sqrt{\mathrm{3}}{a}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\Rightarrow{a}=\sqrt{\frac{\mathrm{4}{A}}{\sqrt{\mathrm{3}}}}=\frac{\mathrm{2}\sqrt{\mathrm{3}\sqrt{\mathrm{3}}{A}}}{\mathrm{3}} \\ $$$${a}_{{red}} =\frac{\mathrm{2}\sqrt{\mathrm{3}\sqrt{\mathrm{3}}×\mathrm{32}}}{\mathrm{3}}=\frac{\mathrm{8}\sqrt{\mathrm{6}\sqrt{\mathrm{3}}}}{\mathrm{3}} \\ $$$${a}_{{blue}} =\frac{\mathrm{2}\sqrt{\mathrm{3}\sqrt{\mathrm{3}}×\mathrm{6}}}{\mathrm{3}}=\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{3}}} \\ $$$${let}\:\delta={PM} \\ $$$${a}_{{red}} −{a}_{{blue}} =\mathrm{2}\sqrt{\mathrm{3}}\delta \\ $$$$\frac{\mathrm{8}\sqrt{\mathrm{6}\sqrt{\mathrm{3}}}}{\mathrm{3}}−\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{3}}}=\mathrm{2}\sqrt{\mathrm{3}}\delta \\ $$$$\Rightarrow\delta=\frac{\mathrm{4}\sqrt{\mathrm{2}\sqrt{\mathrm{3}}}−\sqrt{\mathrm{6}\sqrt{\mathrm{3}}}}{\mathrm{3}}\approx\mathrm{1}.\mathrm{407} \\ $$

Commented by behi83417@gmail.com last updated on 07/Dec/18

nice work sir.thanks in advance.

$${nice}\:{work}\:{sir}.{thanks}\:{in}\:{advance}. \\ $$

Commented by mr W last updated on 07/Dec/18

thank you sir for checking it! it is fixed now.

$${thank}\:{you}\:{sir}\:{for}\:{checking}\:{it}!\:{it} \\ $$$${is}\:{fixed}\:{now}. \\ $$

Answered by behi83417@gmail.com last updated on 07/Dec/18

a_(red) ^2 =(4/(√3)).32=((128)/(√3))⇒a_(red) =((8(√2))/(3)^(1/4) ) a_(blue) ^2 =(4/(√3)).6=((24)/(√3))⇒a_(blue) =((2(√6))/(3)^(1/4) ) ((PM)/(sin30^• ))+h_(blue) +PM=h_(red) PM=((h_(red) −h_(blue) )/((1+sin30)/(sin30)))=((((√3)/2)(a_(red) −a_(blue) ))/3)= =((√3)/6)(((8(√2))/(3)^(1/4) )−((2(√6))/(3)^(1/4) ))=((3)^(1/4) /3)(4(√2)−(√6))#1.4070 .

$${a}_{{red}} ^{\mathrm{2}} =\frac{\mathrm{4}}{\sqrt{\mathrm{3}}}.\mathrm{32}=\frac{\mathrm{128}}{\sqrt{\mathrm{3}}}\Rightarrow{a}_{{red}} =\frac{\mathrm{8}\sqrt{\mathrm{2}}}{\sqrt[{\mathrm{4}}]{\mathrm{3}}} \\ $$$${a}_{{blue}} ^{\mathrm{2}} =\frac{\mathrm{4}}{\sqrt{\mathrm{3}}}.\mathrm{6}=\frac{\mathrm{24}}{\sqrt{\mathrm{3}}}\Rightarrow{a}_{{blue}} =\frac{\mathrm{2}\sqrt{\mathrm{6}}}{\sqrt[{\mathrm{4}}]{\mathrm{3}}} \\ $$$$\frac{{PM}}{{sin}\mathrm{30}^{\bullet} }+{h}_{{blue}} +{PM}={h}_{{red}} \\ $$$${PM}=\frac{{h}_{{red}} −{h}_{{blue}} }{\frac{\mathrm{1}+{sin}\mathrm{30}}{{sin}\mathrm{30}}}=\frac{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left({a}_{{red}} −{a}_{{blue}} \right)}{\mathrm{3}}= \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{6}}\left(\frac{\mathrm{8}\sqrt{\mathrm{2}}}{\sqrt[{\mathrm{4}}]{\mathrm{3}}}−\frac{\mathrm{2}\sqrt{\mathrm{6}}}{\sqrt[{\mathrm{4}}]{\mathrm{3}}}\right)=\frac{\sqrt[{\mathrm{4}}]{\mathrm{3}}}{\mathrm{3}}\left(\mathrm{4}\sqrt{\mathrm{2}}−\sqrt{\mathrm{6}}\right)#\mathrm{1}.\mathrm{4070}\:. \\ $$

Commented by Kunal12588 last updated on 07/Dec/18

sir what is the meaning of #

Commented by behi83417@gmail.com last updated on 07/Dec/18

i use this sign to say: approximately or almost or nearly....

$${i}\:{use}\:{this}\:{sign}\:{to}\:{say}:\:{approximately} \\ $$$${or}\:{almost}\:{or}\:{nearly}.... \\ $$

Answered by Kunal12588 last updated on 08/Dec/18

a_(blue) =(√(((24)/(√3)) )), a_(red) =(√((128)/(√3))) join T & A and S & B , a_(red) ∥a_(blue) so quad.ABST is a trapezium ar(ABST)=(1/3){ar(△)−ar(△)}=(1/3)(32−6) (1/2)(a_(blue) +a_(red) )PM= ((26)/3) PM{(√(((24)/(√3)) ))+(√((128)/(√3)))}=((52)/3) PM=((52)/(3((√(((24)/(√3)) ))+(√((128)/(√3))))))=1.40704302...

$${a}_{{blue}} =\sqrt{\frac{\mathrm{24}}{\sqrt{\mathrm{3}}}\:},\:{a}_{{red}} =\sqrt{\frac{\mathrm{128}}{\sqrt{\mathrm{3}}}} \\ $$$${join}\:{T}\:\&\:{A}\:{and}\:{S}\:\&\:{B}\:,\:{a}_{{red}} \parallel{a}_{{blue}} \\ $$$${so}\:{quad}.{ABST}\:{is}\:{a}\:{trapezium} \\ $$$${ar}\left({ABST}\right)=\frac{\mathrm{1}}{\mathrm{3}}\left\{{ar}\left(\bigtriangleup\right)−{ar}\left(\bigtriangleup\right)\right\}=\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{32}−\mathrm{6}\right) \\ $$$$\:\frac{\mathrm{1}}{\mathrm{2}}\left({a}_{{blue}} +{a}_{{red}} \right){PM}=\:\frac{\mathrm{26}}{\mathrm{3}} \\ $$$${PM}\left\{\sqrt{\frac{\mathrm{24}}{\sqrt{\mathrm{3}}}\:}+\sqrt{\frac{\mathrm{128}}{\sqrt{\mathrm{3}}}}\right\}=\frac{\mathrm{52}}{\mathrm{3}} \\ $$$${PM}=\frac{\mathrm{52}}{\mathrm{3}\left(\sqrt{\frac{\mathrm{24}}{\sqrt{\mathrm{3}}}\:}+\sqrt{\frac{\mathrm{128}}{\sqrt{\mathrm{3}}}}\right)}=\mathrm{1}.\mathrm{40704302}... \\ $$

Commented by Kunal12588 last updated on 08/Dec/18

thanks for checking sir

$${thanks}\:{for}\:{checking}\:{sir} \\ $$

Commented by behi83417@gmail.com last updated on 07/Dec/18

thank you sir.beautiful method! ((54)/3)⇒((52)/3) (replace please)

$${thank}\:{you}\:{sir}.{beautiful}\:{method}! \\ $$$$\frac{\mathrm{54}}{\mathrm{3}}\Rightarrow\frac{\mathrm{52}}{\mathrm{3}}\:\left({replace}\:{please}\right) \\ $$

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