Menu Close

Question-49536




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

Leave a Reply

Your email address will not be published. Required fields are marked *