Question-104841

Question Number 104841 by bramlex last updated on 24/Jul/20

Commented by 1549442205PVT last updated on 24/Jul/20

please,who can translate this language to English or tell me it is what language so i can get help from other person.

$$\mathrm{please},\mathrm{who}\:\mathrm{can}\:\mathrm{translate}\:\mathrm{this} \\ $$$$\mathrm{language}\:\mathrm{to}\:\mathrm{English}\:\mathrm{or}\:\mathrm{tell}\:\mathrm{me}\:\mathrm{it}\:\mathrm{is} \\ $$$$\mathrm{what}\:\:\mathrm{language}\:\mathrm{so}\:\mathrm{i}\:\mathrm{can}\:\mathrm{get}\:\mathrm{help}\:\mathrm{from} \\ $$$$\mathrm{other}\:\mathrm{person}. \\ $$

Commented by $@y@m last updated on 24/Jul/20

$${I}\:{think}\:{translation}\:{is}\:{not}\:{required}. \\ $$$${I}\:{don}'{t}\:{know}\:{this}\:{language}\:{literally}, \\ $$$${but}\:{understood}\:{the}\:{question}\:{and} \\ $$$${solved}\:{it}\:{because}\:{it}\:{has}\:{universal} \\ $$$${language}\:{of}\:{Math}. \\ $$

Commented by 1549442205PVT last updated on 24/Jul/20

That seems from Uzbek language .Don′t know what is question of problem.You have a good guess

$$\mathrm{That}\:\mathrm{seems}\:\mathrm{from}\:\mathrm{Uzbek}\:\mathrm{language} \\ $$$$.\mathrm{Don}'\mathrm{t}\:\mathrm{know}\:\:\mathrm{what}\:\mathrm{is}\:\mathrm{question}\:\mathrm{of} \\ $$$$\mathrm{problem}.\mathrm{You}\:\mathrm{have}\:\mathrm{a}\:\mathrm{good}\:\mathrm{guess} \\ $$

Answered by john santu last updated on 24/Jul/20

⇒ ((12.5)/2) + 26 +((ax.sin α)/2) = (((x+5).12)/2) ⇒30 + 26 + ((ax.sin α)/2) = 6x +30 ⇒26 + ((ax.sin α)/2) = 6x...(1) ⇒((a.sin α)/2) = 26 ⇒a.sin α = 4...(2) ⇒ 26 + 2x = 6x ; 13 = 2x ; x = ((13)/2) (JS ⊛ )

$$\Rightarrow\:\frac{\mathrm{12}.\mathrm{5}}{\mathrm{2}}\:+\:\mathrm{26}\:+\frac{{ax}.\mathrm{sin}\:\alpha}{\mathrm{2}}\:=\:\frac{\left({x}+\mathrm{5}\right).\mathrm{12}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{30}\:+\:\mathrm{26}\:+\:\frac{{ax}.\mathrm{sin}\:\alpha}{\mathrm{2}}\:=\:\mathrm{6}{x}\:+\mathrm{30} \\ $$$$\Rightarrow\mathrm{26}\:+\:\frac{{ax}.\mathrm{sin}\:\alpha}{\mathrm{2}}\:=\:\mathrm{6}{x}…\left(\mathrm{1}\right) \\ $$$$\Rightarrow\frac{{a}.\mathrm{sin}\:\alpha}{\mathrm{2}}\:=\:\mathrm{26}\:\Rightarrow{a}.\mathrm{sin}\:\alpha\:=\:\mathrm{4}…\left(\mathrm{2}\right) \\ $$$$\Rightarrow\:\mathrm{26}\:+\:\mathrm{2}{x}\:=\:\mathrm{6}{x}\:;\:\mathrm{13}\:=\:\mathrm{2}{x}\:;\:{x}\:=\:\frac{\mathrm{13}}{\mathrm{2}} \\ $$$$\left({JS}\:\circledast\:\right)\: \\ $$

Commented by bramlex last updated on 24/Jul/20

$${thank}\:{you}\:\clubsuit \\ $$

Answered by $@y@m last updated on 24/Jul/20

S(AED)=26 (1/2)×ED×13sin θ=26 EDsin θ=4 ....(1) S(ABD)=(1/2)×x×12 =6x .....(2) S(EBD)=(1/2)×ED×xsin θ=2x {using (1) Now, S(ABD)=S(AED)+S(EBD) ⇒6x=26+2x 4x=26 x=6,5 sm

$${S}\left({AED}\right)=\mathrm{26} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×{ED}×\mathrm{13sin}\:\theta=\mathrm{26} \\ $$$${ED}\mathrm{sin}\:\theta=\mathrm{4}\:….\left(\mathrm{1}\right) \\ $$$${S}\left({ABD}\right)=\frac{\mathrm{1}}{\mathrm{2}}×{x}×\mathrm{12}\:=\mathrm{6}{x}\:…..\left(\mathrm{2}\right) \\ $$$${S}\left({EBD}\right)=\frac{\mathrm{1}}{\mathrm{2}}×{ED}×{x}\mathrm{sin}\:\theta=\mathrm{2}{x}\:\left\{{using}\:\left(\mathrm{1}\right)\right. \\ $$$${Now}, \\ $$$${S}\left({ABD}\right)={S}\left({AED}\right)+{S}\left({EBD}\right) \\ $$$$\Rightarrow\mathrm{6}{x}=\mathrm{26}+\mathrm{2}{x} \\ $$$$\mathrm{4}{x}=\mathrm{26} \\ $$$${x}=\mathrm{6},\mathrm{5}\:{sm} \\ $$

Commented by bramlex last updated on 24/Jul/20

$${thank}\:{you}\:\spadesuit \\ $$

Answered by 1549442205PVT last updated on 24/Jul/20

If question of problem being find the length x of segment BD: AC=(√(AD^2 −CD^2 ))=(√(13^2 −5^2 ))=(√(144))=12cm^2 S(ABD)=(1/2)BD×AC=(1/2).x.12=6x(cm^2 ) DE is the bisector so ((AE)/(EB))=((AD)/(BD))=((13)/x) ((S(AED))/(S(BED)))=((2S(AED))/(2S(BED)))=((DA.DEsinθ)/(DB.DEsinθ))=((AD)/(BD))=((13)/x) ⇒((26)/(6x−26))=((13)/x)⇔26x=78x−338 ⇒52x=338⇔x=((338)/(52))=6,5 cm Hence,we choose answer D(6,5)

$$\mathrm{If}\:\mathrm{question}\:\mathrm{of}\:\mathrm{problem}\:\mathrm{being}\:\mathrm{find}\:\mathrm{the}\:\mathrm{length}\:\mathrm{x}\:\mathrm{of}\:\mathrm{segment}\:\mathrm{BD}: \\ $$$$\mathrm{AC}=\sqrt{\mathrm{AD}^{\mathrm{2}} −\mathrm{CD}^{\mathrm{2}} }=\sqrt{\mathrm{13}^{\mathrm{2}} −\mathrm{5}^{\mathrm{2}} }=\sqrt{\mathrm{144}}=\mathrm{12cm}^{\mathrm{2}} \\ $$$$\mathrm{S}\left(\mathrm{ABD}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{BD}×\mathrm{AC}=\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{x}.\mathrm{12}=\mathrm{6x}\left(\mathrm{cm}^{\mathrm{2}} \right) \\ $$$$\mathrm{DE}\:\mathrm{is}\:\mathrm{the}\:\mathrm{bisector}\:\mathrm{so}\:\:\frac{\mathrm{AE}}{\mathrm{EB}}=\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{13}}{\mathrm{x}} \\ $$$$\frac{\mathrm{S}\left(\mathrm{AED}\right)}{\mathrm{S}\left(\mathrm{BED}\right)}=\frac{\mathrm{2S}\left(\mathrm{AED}\right)}{\mathrm{2S}\left(\mathrm{BED}\right)}=\frac{\mathrm{DA}.\mathrm{DEsin}\theta}{\mathrm{DB}.\mathrm{DEsin}\theta}=\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{13}}{\mathrm{x}} \\ $$$$\Rightarrow\frac{\mathrm{26}}{\mathrm{6x}−\mathrm{26}}=\frac{\mathrm{13}}{\mathrm{x}}\Leftrightarrow\mathrm{26x}=\mathrm{78x}−\mathrm{338} \\ $$$$\Rightarrow\mathrm{52x}=\mathrm{338}\Leftrightarrow\mathrm{x}=\frac{\mathrm{338}}{\mathrm{52}}=\mathrm{6},\mathrm{5}\:\mathrm{cm} \\ $$$$\boldsymbol{\mathrm{Hence}},\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{choose}}\:\boldsymbol{\mathrm{answer}}\:\boldsymbol{\mathrm{D}}\left(\mathrm{6},\mathrm{5}\right) \\ $$

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