Đáp án:
$V_{AMNP}=7a^3$
Giải thích các bước giải:
$V_{ABCD}=\dfrac13.AD.\dfrac12.AB.AC=\dfrac16.AB.AC.AD=\dfrac16.6a.7a.4a=28a^3$
Áp dụng công thức tính tỉ lệ thể tích các khối tứ diện ta có:
$\dfrac{V_{DAPN}}{V_{DABC}}=\dfrac{DA}{DA}.\dfrac{DP}{DB}.\dfrac{DN}{DC}=\dfrac12.\dfrac12=\dfrac14$
$\Rightarrow V_{DAPN}=\dfrac14.V_{DABC}$
$\dfrac{V_{BAPM}}{V_{BADC}}=\dfrac{BA}{BA}.\dfrac{BP}{BD}.\dfrac{BM}{BC}=\dfrac12.\dfrac12=\dfrac14$
$\Rightarrow V_{BAPM}=\dfrac14.V_{BADC}$
$\dfrac{V_{CAMN}}{V_{CABD}}=\dfrac{CA}{CA}.\dfrac{CM}{CB}.\dfrac{CN}{CD}=\dfrac12.\dfrac12=\dfrac14$
$\Rightarrow V_{CAMN}=\dfrac14.V_{CABD}$
$\Rightarrow V_{APMN}=V_{ABCD}-V_{DAPN}-V_{BAPM}-V_{CAMN}$
$=V_{ABCD}-\dfrac14V_{ABCD}-\dfrac14V_{ABCD}-\dfrac14V_{ABCD}$
$=V_{ABCD}-\dfrac34V_{ABCD}=\dfrac14V_{ABCD}=\dfrac{28a^3}4=7a^3$
