Question Number 15318 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 09/Jun/17
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 09/Jun/17
$${given}:\:{ABCD},{rectangle}.{connect}\: \\ $$$${each}\:{vertex}\:{to}\:{the}\:{middle}\:{point}\:{of} \\ $$$${one}\:{other}\:{sides}\:{as}\:{shown}. \\ $$$$\left.\mathrm{1}\right){show}\:{that}\:{MNPQ},{is}\:{a}\:{parallelogram} \\ $$$${and}\:{its}\:{area}\:{is}\:{equail}\:{to}:\:\frac{\mathrm{1}}{\mathrm{5}}{area}\:{of}\:{ABCD}. \\ $$$$\left.\mathrm{2}\right){S}_{{S}\overset{\Delta} {{P}C}} ={S}_{{B}\overset{\Delta} {{N}T}} =\frac{\mathrm{1}}{\mathrm{20}}.{S}_{{ABCD}} . \\ $$
Commented by mrW1 last updated on 09/Jun/17
Commented by mrW1 last updated on 09/Jun/17
$$\mathrm{S}_{\mathrm{1}} =\mathrm{S}_{\mathrm{2}} =\mathrm{S}_{\mathrm{3}} =\mathrm{S}_{\mathrm{4}} =\mathrm{S}_{\mathrm{5}} \\ $$$$\mathrm{S}_{\mathrm{1}} +\mathrm{S}_{\mathrm{2}} +\mathrm{S}_{\mathrm{3}} +\mathrm{S}_{\mathrm{4}} +\mathrm{S}_{\mathrm{5}} =\mathrm{S}_{\mathrm{ABCD}} \\ $$$$\Rightarrow\mathrm{S}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{5}}\mathrm{S}_{\mathrm{ABCD}} \\ $$$$ \\ $$$$\mathrm{S}_{\Delta\mathrm{SPC}} =\frac{\mathrm{1}}{\mathrm{4}}\mathrm{S}_{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{1}}{\mathrm{5}}\mathrm{S}_{\mathrm{ABCD}} =\frac{\mathrm{1}}{\mathrm{20}}\mathrm{S}_{\mathrm{ABCD}} \\ $$$$\mathrm{S}_{\Delta\mathrm{BNT}} =\frac{\mathrm{1}}{\mathrm{4}}\mathrm{S}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{1}}{\mathrm{5}}\mathrm{S}_{\mathrm{ABCD}} =\frac{\mathrm{1}}{\mathrm{20}}\mathrm{S}_{\mathrm{ABCD}} \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 09/Jun/17
$${thank}\:{you}\:{master}.{it}\:{is}\:{simple}\:{and}\:{perfect}. \\ $$
Commented by mrW1 last updated on 09/Jun/17
$$\mathrm{what}\:\mathrm{if}\:\mathrm{not}\:\mathrm{the}\:\mathrm{middle}\:\mathrm{point}\:\mathrm{but}\:\mathrm{the} \\ $$$$\mathrm{1}/\mathrm{3}−\mathrm{point}\:\mathrm{is}\:\mathrm{connected}\:\mathrm{with}\:\mathrm{the} \\ $$$$\mathrm{vertex}? \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 10/Jun/17
$${in}\:{case}\:{of}\::\frac{\mathrm{1}}{\mathrm{3}}\:,\frac{{S}_{{ABCD}} }{{S}_{{MNPQ}} }=\mathrm{2}.\mathrm{5},\frac{{S}_{{ABCD}} }{{S}_{{BNT}} }=\mathrm{60}. \\ $$
Answered by ajfour last updated on 10/Jun/17
Commented by ajfour last updated on 10/Jun/17
$${all}\://{gms}\:{that}\:{seem}\:{like}\:{the} \\ $$$${central}\:{one}\:{are}\:{alike},\:{which}\: \\ $$$${is}\:{not}\:{so}\:{in}\:{diagram},\:{but}\:{can} \\ $$$${be}\:{judged}\:{so}. \\ $$$$\:\mathrm{Then}\:\mathrm{5}\:\mathrm{such}\://\mathrm{gm}\:\mathrm{areas}= \\ $$$$\:\:\:\mathrm{A}{rea}\:{of}\:{rectangle} \\ $$$${The}\:{corner}\:\Delta'{s}\:{have}\:{one}-{fourth} \\ $$$${of}\:{the}\://{gm}\:{area}=\:{one}-{twentieth} \\ $$$${of}\:{the}\:{rectangle}\:{area}. \\ $$
Commented by ajfour last updated on 10/Jun/17
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 10/Jun/17
$${tg}\angle{NBT}={tg}\angle{ADM}=\frac{{b}}{\mathrm{2}{a}}\Rightarrow{QM}\parallel{PN}\left(\mathrm{1}\right) \\ $$$${tg}\angle{QCD}={tg}\angle{TAB}=\frac{{a}}{\mathrm{2}{b}}\Rightarrow{PQ}\parallel{MN}\left(\mathrm{2}\right) \\ $$$${from}\left(\mathrm{1}\right),\left(\mathrm{2}\right)\Rightarrow{MNPQ},{is}\:{a}\:{parallelogram}. \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 10/Jun/17
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 10/Jun/17
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 10/Jun/17
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 10/Jun/17
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 10/Jun/17
