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Question-83694




Question Number 83694 by naka3546 last updated on 05/Mar/20
Commented by naka3546 last updated on 05/Mar/20
DE = EF = FC ,  BG = 2 CG ,  ABCD  is  a  square .  (([ BMN ])/([ ABCD ]))  =  ?  ( without  Trigonometry  or Menelause )
$${DE}\:=\:{EF}\:=\:{FC}\:, \\ $$$${BG}\:=\:\mathrm{2}\:{CG}\:, \\ $$$${ABCD}\:\:{is}\:\:{a}\:\:{square}\:. \\ $$$$\frac{\left[\:{BMN}\:\right]}{\left[\:{ABCD}\:\right]}\:\:=\:\:? \\ $$$$\left(\:{without}\:\:{Trigonometry}\:\:{or}\:{Menelause}\:\right) \\ $$
Commented by john santu last updated on 05/Mar/20
what the meaning [ BMN] ?   area ?
$$\mathrm{what}\:\mathrm{the}\:\mathrm{meaning}\:\left[\:\mathrm{BMN}\right]\:?\: \\ $$$$\mathrm{area}\:?\: \\ $$
Answered by MJS last updated on 05/Mar/20
A= ((0),(3) ) B= ((3),(3) ) C= ((3),(0) ) D= ((0),(0) )  E= ((1),(0) ) F= ((2),(0) ) G= ((3),(1) )  line DG: y=(1/3)x  line EB: y=(3/2)x−(3/2)  line FB: y=3x−6  DG∩EB=M= (((9/7)),((3/7)) )  DG∩FB=N= (((9/4)),((3/4)) )  ∣BM∣=((6(√(13)))/7)  ∣BN∣=((3(√(10)))/4)  ∣MN∣=((9(√(10)))/(28))  area (abc) =(1/4)(√((a+b+c)(−a+b+c)(a+b+c)(a+b−c)))  ⇒ area (BMN) =((27)/(28))  area (ABCD)=9  ⇒ ((area (BMN))/(area (ABCD)))=(3/(28))
$${A}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{3}}\end{pmatrix}\:{B}=\begin{pmatrix}{\mathrm{3}}\\{\mathrm{3}}\end{pmatrix}\:{C}=\begin{pmatrix}{\mathrm{3}}\\{\mathrm{0}}\end{pmatrix}\:{D}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix} \\ $$$${E}=\begin{pmatrix}{\mathrm{1}}\\{\mathrm{0}}\end{pmatrix}\:{F}=\begin{pmatrix}{\mathrm{2}}\\{\mathrm{0}}\end{pmatrix}\:{G}=\begin{pmatrix}{\mathrm{3}}\\{\mathrm{1}}\end{pmatrix} \\ $$$$\mathrm{line}\:{DG}:\:{y}=\frac{\mathrm{1}}{\mathrm{3}}{x} \\ $$$$\mathrm{line}\:{EB}:\:{y}=\frac{\mathrm{3}}{\mathrm{2}}{x}−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{line}\:{FB}:\:{y}=\mathrm{3}{x}−\mathrm{6} \\ $$$${DG}\cap{EB}={M}=\begin{pmatrix}{\mathrm{9}/\mathrm{7}}\\{\mathrm{3}/\mathrm{7}}\end{pmatrix} \\ $$$${DG}\cap{FB}={N}=\begin{pmatrix}{\mathrm{9}/\mathrm{4}}\\{\mathrm{3}/\mathrm{4}}\end{pmatrix} \\ $$$$\mid{BM}\mid=\frac{\mathrm{6}\sqrt{\mathrm{13}}}{\mathrm{7}} \\ $$$$\mid{BN}\mid=\frac{\mathrm{3}\sqrt{\mathrm{10}}}{\mathrm{4}} \\ $$$$\mid{MN}\mid=\frac{\mathrm{9}\sqrt{\mathrm{10}}}{\mathrm{28}} \\ $$$$\mathrm{area}\:\left({abc}\right)\:=\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}+{b}+{c}\right)\left({a}+{b}−{c}\right)} \\ $$$$\Rightarrow\:\mathrm{area}\:\left({BMN}\right)\:=\frac{\mathrm{27}}{\mathrm{28}} \\ $$$$\mathrm{area}\:\left({ABCD}\right)=\mathrm{9} \\ $$$$\Rightarrow\:\frac{\mathrm{area}\:\left({BMN}\right)}{\mathrm{area}\:\left({ABCD}\right)}=\frac{\mathrm{3}}{\mathrm{28}} \\ $$
Answered by mr W last updated on 05/Mar/20
Commented by mr W last updated on 05/Mar/20
ABCD is rectangle, mustn′t be square.  ((EM)/(MB))=((PE)/(BG))=(1/3)×((CG)/(BG))=(1/3)×(1/2)=(1/6)  ⇒((MB)/(EB))=(6/7)  ((FN)/(NB))=((QF)/(BG))=(2/3)×((CG)/(BG))=(2/3)×(1/2)=(1/3)  ⇒((NB)/(FB))=(3/4)  (([BMN])/([BEF]))=((MB)/(EB))×((NB)/(FB))=(6/7)×(3/4)=((18)/(28))  (([BEF])/([ABCD]))=(1/6)  ⇒(([BMN])/([ABCD]))=((18)/(28))×(1/6)=(3/(28))
$${ABCD}\:{is}\:{rectangle},\:{mustn}'{t}\:{be}\:{square}. \\ $$$$\frac{{EM}}{{MB}}=\frac{{PE}}{{BG}}=\frac{\mathrm{1}}{\mathrm{3}}×\frac{{CG}}{{BG}}=\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\Rightarrow\frac{{MB}}{{EB}}=\frac{\mathrm{6}}{\mathrm{7}} \\ $$$$\frac{{FN}}{{NB}}=\frac{{QF}}{{BG}}=\frac{\mathrm{2}}{\mathrm{3}}×\frac{{CG}}{{BG}}=\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow\frac{{NB}}{{FB}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\frac{\left[{BMN}\right]}{\left[{BEF}\right]}=\frac{{MB}}{{EB}}×\frac{{NB}}{{FB}}=\frac{\mathrm{6}}{\mathrm{7}}×\frac{\mathrm{3}}{\mathrm{4}}=\frac{\mathrm{18}}{\mathrm{28}} \\ $$$$\frac{\left[{BEF}\right]}{\left[{ABCD}\right]}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\Rightarrow\frac{\left[{BMN}\right]}{\left[{ABCD}\right]}=\frac{\mathrm{18}}{\mathrm{28}}×\frac{\mathrm{1}}{\mathrm{6}}=\frac{\mathrm{3}}{\mathrm{28}} \\ $$

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