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 )](https://www.tinkutara.com/question/Q83695.png)
$${DE}\:=\:{EF}\:=\:{FC}\:, \\ $$$${BG}\:=\:\mathrm{2}\:{CG}\:, \\ $$$${ABCD}\:\:{is}\:\:{a}\:\:{square}\:. \\ $$$$\frac{\left[\:{BMN}\:\right]}{\left[\:{ABCD}\:\right]}\:\:=\:\:? \\ $$$$\left(\:{without}\:\:{Trigonometry}\:\:{or}\:{Menelause}\:\right) \\ $$
Commented by john santu last updated on 05/Mar/20
![what the meaning [ BMN] ? area ?](https://www.tinkutara.com/question/Q83699.png)
$$\mathrm{what}\:\mathrm{the}\:\mathrm{meaning}\:\left[\:\mathrm{BMN}\right]\:?\: \\ $$$$\mathrm{area}\:?\: \\ $$
Answered by MJS last updated on 05/Mar/20
$${A}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{3}}\end{pmatrix}\:{B}=\begin{pmatrix}{\mathrm{3}}\\{\mathrm{3}}\end{pmatrix}\:{C}=\begin{pmatrix}{\mathrm{3}}\\{\mathrm{0}}\end{pmatrix}\:{D}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix} \\ $$$${E}=\begin{pmatrix}{\mathrm{1}}\\{\mathrm{0}}\end{pmatrix}\:{F}=\begin{pmatrix}{\mathrm{2}}\\{\mathrm{0}}\end{pmatrix}\:{G}=\begin{pmatrix}{\mathrm{3}}\\{\mathrm{1}}\end{pmatrix} \\ $$$$\mathrm{line}\:{DG}:\:{y}=\frac{\mathrm{1}}{\mathrm{3}}{x} \\ $$$$\mathrm{line}\:{EB}:\:{y}=\frac{\mathrm{3}}{\mathrm{2}}{x}−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{line}\:{FB}:\:{y}=\mathrm{3}{x}−\mathrm{6} \\ $$$${DG}\cap{EB}={M}=\begin{pmatrix}{\mathrm{9}/\mathrm{7}}\\{\mathrm{3}/\mathrm{7}}\end{pmatrix} \\ $$$${DG}\cap{FB}={N}=\begin{pmatrix}{\mathrm{9}/\mathrm{4}}\\{\mathrm{3}/\mathrm{4}}\end{pmatrix} \\ $$$$\mid{BM}\mid=\frac{\mathrm{6}\sqrt{\mathrm{13}}}{\mathrm{7}} \\ $$$$\mid{BN}\mid=\frac{\mathrm{3}\sqrt{\mathrm{10}}}{\mathrm{4}} \\ $$$$\mid{MN}\mid=\frac{\mathrm{9}\sqrt{\mathrm{10}}}{\mathrm{28}} \\ $$$$\mathrm{area}\:\left({abc}\right)\:=\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}+{b}+{c}\right)\left({a}+{b}−{c}\right)} \\ $$$$\Rightarrow\:\mathrm{area}\:\left({BMN}\right)\:=\frac{\mathrm{27}}{\mathrm{28}} \\ $$$$\mathrm{area}\:\left({ABCD}\right)=\mathrm{9} \\ $$$$\Rightarrow\:\frac{\mathrm{area}\:\left({BMN}\right)}{\mathrm{area}\:\left({ABCD}\right)}=\frac{\mathrm{3}}{\mathrm{28}} \\ $$
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))](https://www.tinkutara.com/question/Q83709.png)
$${ABCD}\:{is}\:{rectangle},\:{mustn}'{t}\:{be}\:{square}. \\ $$$$\frac{{EM}}{{MB}}=\frac{{PE}}{{BG}}=\frac{\mathrm{1}}{\mathrm{3}}×\frac{{CG}}{{BG}}=\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\Rightarrow\frac{{MB}}{{EB}}=\frac{\mathrm{6}}{\mathrm{7}} \\ $$$$\frac{{FN}}{{NB}}=\frac{{QF}}{{BG}}=\frac{\mathrm{2}}{\mathrm{3}}×\frac{{CG}}{{BG}}=\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow\frac{{NB}}{{FB}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\frac{\left[{BMN}\right]}{\left[{BEF}\right]}=\frac{{MB}}{{EB}}×\frac{{NB}}{{FB}}=\frac{\mathrm{6}}{\mathrm{7}}×\frac{\mathrm{3}}{\mathrm{4}}=\frac{\mathrm{18}}{\mathrm{28}} \\ $$$$\frac{\left[{BEF}\right]}{\left[{ABCD}\right]}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\Rightarrow\frac{\left[{BMN}\right]}{\left[{ABCD}\right]}=\frac{\mathrm{18}}{\mathrm{28}}×\frac{\mathrm{1}}{\mathrm{6}}=\frac{\mathrm{3}}{\mathrm{28}} \\ $$