Cho tứ diện ABCD có AB=CD=11m, BC=AD=20m, BD=AC=21m Tính thể tích khối tứ diện ABCD.

Đáp án D

Suy ra  $V_{PADB}+V_{MABC}+V_{QBCD}+V_{NACD}=\frac{1}{6}V+\frac{1}{6}V+\frac{1}{6}V+\frac{1}{6}V=\frac{2}{3}V.$

Mà  $V_{PADB}+V_{MABC}+V_{QBCD}+V_{NACD}+V_{ABCD}=V$

Suy ra:  $V_{ABCD}=\frac{1}{3}V=m.n.p$

Xét các tam giác vuông APB; APD; PDB, theo định lý Pytago ta có:

$\{\begin{array}{l}{m}^2+n^2=B{D}^2\\ m^2+p^2=A{D}^2\\ p^2+n^2=A{B}^2\end{array}\iff \{\begin{array}{l}{m}^2+n^2={21}^2\\ m^2+p^2={20}^2\\ p^2+n^2={11}^2\end{array}\iff \{\begin{array}{l}{m}^2+n^2+p^2=481\\ m^2+n^2={21}^2\\ m^2+p^2={20}^2\\ p^2+n^2={11}^2\end{array}\iff \{\begin{array}{l}m=6\sqrt{10}\\ n=9\\ p=2\sqrt{10}\end{array}$ 

$V_{ABCD}=\frac{1}{3}m.n.p=\frac{1}{3}.6\sqrt{10}\mathrm{.9.2}\sqrt{10}=360m^3$