Đáp án: $\dfrac{{{S}_{MNPQ}}}{{{S}_{ABCD}}}=\dfrac{1}{2}$
Giải thích:
Phân tích:
${{S}_{MNPQ}}={{S}_{ABCD}}-\left( {{S}_{AMQ}}+{{S}_{CNP}}+{{S}_{DQP}}+{{S}_{BMN}} \right)$
$MQ$ là đường trung bình $\Delta ABD$
$\to MQ\,\,||\,\,BD$
$\to \Delta AMQ\backsim\Delta ABD$
$\to \dfrac{{{S}_{AMQ}}}{{{S}_{ABD}}}={{\left( \dfrac{AM}{AB} \right)}^{2}}={{\left( \dfrac{1}{2} \right)}^{2}}=\dfrac{1}{4}$
$\to {{S}_{AMQ}}=\dfrac{1}{4}{{S}_{ABD}}\,\,\,\,\,\left( 1 \right)$
$NP$ là đường trung bình $\Delta CBD$
$\to NP\,\,||\,\,BD$
$\to \Delta CNP\backsim\Delta CBD$
$\to \dfrac{{{S}_{CNP}}}{{{S}_{CBD}}}={{\left( \dfrac{CN}{CB} \right)}^{2}}={{\left( \dfrac{1}{2} \right)}^{2}}=\dfrac{1}{4}$
$\to {{S}_{CNP}}=\dfrac{1}{4}{{S}_{CBD}}\,\,\,\,\,\left( 2 \right)$
Lấy $\left( 1 \right)+\left( 2 \right)$, cộng vế theo vế, ta được:
${{S}_{AMQ}}+{{S}_{CNP}}=\dfrac{1}{4}\left( {{S}_{ABD}}+{{S}_{CBD}} \right)=\dfrac{1}{4}{{S}_{ABCD}}$
Chứng minh hoàn toàn tương tự:
${{S}_{DQP}}+{{S}_{BMN}}=\dfrac{1}{4}\left( {{S}_{DAC}}+{{S}_{BAC}} \right)=\dfrac{1}{4}{{S}_{ABCD}}$
$\,\,\,\,\,\,\,{{S}_{MNPQ}}={{S}_{ABCD}}-\left( {{S}_{AMQ}}+{{S}_{CNP}}+{{S}_{DQP}}+{{S}_{BMN}} \right)$
$\to {{S}_{MNPQ}}={{S}_{ABCD}}-\left( \dfrac{1}{4}{{S}_{ABCD}}+\dfrac{1}{4}{{S}_{ABCD}} \right)$
$\to {{S}_{MNPQ}}=\dfrac{1}{2}{{S}_{ABCD}}$
$\to \dfrac{{{S}_{MNPQ}}}{{{S}_{ABCD}}}=\dfrac{1}{2}$
