Bài tập 8 trang 68 SBT Toán 11 Tập 2 Chân trời sáng tạo - CTST

Ta có:

$V_{ABCD.A\text{'}B\text{'}C\text{'}D\text{'}}=V_{BAB\text{'}C}+V_{D.ACD\text{'}}+V_{A\text{'}B\text{'}AD\text{'}}+V_{C\text{'}B\text{'}CD\text{'}}+V_{ACB\text{'}D\text{'}}=4.V_{BAB\text{'}C}+V_{ACB\text{'}D\text{'}}$

$\Rightarrow V_{ACB\text{'}D\text{'}}=V_{ABCD.A\text{'}B\text{'}C\text{'}D\text{'}}-4.V_{BAB\text{'}C}=V_{ABCD.A\text{'}B\text{'}C\text{'}D\text{'}}-4.\frac{1}{6}.V_{ABCD.A\text{'}B\text{'}C\text{'}D\text{'}}=\frac{1}{3}{V}_{ABCD.A\text{'}B\text{'}C\text{'}D\text{'}}$

$\Rightarrow V_{ACB\text{'}D\text{'}}=\frac{1}{3}\mathrm{.2.3.6}=12\left(c{m}^3\right)$

-- Mod Toán 11 HỌC247