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Câu hỏi: Cho hình chóp $S.ABC$ có đáy $ABC$ là tam giác đều cạnh bằng 1. Biết khoảng cách từ $A$ đến mặt phẳng $\left( SBC \right)$ là $\dfrac{\sqrt{6}}{4}$, từ $B$ đến mặt phẳng $\left( SAC \right)$ là $\dfrac{\sqrt{15}}{10}$, từ $C$ đến mặt phẳng $\left( SAB \right)$ là $\dfrac{\sqrt{30}}{20}$.và hình chiếu vuông góc của $S$ xuống đáy nằm trong tam giác $ABC$. Thể tích khối chóp $S.ABC$ bằng
A. $\dfrac{1}{36}$.
B. $\dfrac{1}{48}$.
C. $\dfrac{1}{12}$.
D. $\dfrac{1}{24}$.
Gọi $O$ là chân đường cao hạ từ $S$ xuống mặt phẳng $\left( ABC \right)$
Đặt $d\left( A,BC \right)=a,d\left( B,AC \right)=b,d\left( C,AB \right)=c,SO=h$
Ta có ${{S}_{\Delta ABC}}={{S}_{\Delta OBC}}+{{S}_{\Delta OAC}}+{{S}_{\Delta OAB}}\Rightarrow a+b+c=\dfrac{\sqrt{3}}{2}\left( 1 \right)$
Mặt khác $\dfrac{d\left( O,\left( SBC \right) \right)}{d\left( A,\left( SBC \right) \right)}=\dfrac{OM}{AM}=\dfrac{OI}{AK}=\dfrac{2a}{\sqrt{3}}\Rightarrow d\left( O,\left( SBC \right) \right)=\dfrac{2a}{\sqrt{3}}.\dfrac{\sqrt{6}}{4}=\dfrac{a}{\sqrt{2}}$
$\Rightarrow \dfrac{2}{{{a}^{2}}}=\dfrac{1}{{{h}^{2}}}+\dfrac{1}{{{a}^{2}}}\Rightarrow a=h$
Tương tự $\dfrac{d\left( O,\left( SAC \right) \right)}{d\left( B,\left( SAC \right) \right)}=\dfrac{d\left( O,AC \right)}{d\left( B,AC \right)}==\dfrac{2b}{\sqrt{3}}\Rightarrow d\left( O,\left( SAC \right) \right)=\dfrac{2b}{\sqrt{3}}.\dfrac{\sqrt{15}}{10}=\dfrac{b}{\sqrt{5}}$
$\Rightarrow \dfrac{5}{{{b}^{2}}}=\dfrac{1}{{{h}^{2}}}+\dfrac{1}{{{b}^{2}}}\Rightarrow b=2h$
Tương tự $\dfrac{d\left( O,\left( SAB \right) \right)}{d\left( C,\left( SAB \right) \right)}=\dfrac{d\left( O,AB \right)}{d\left( C,AB \right)}==\dfrac{2c}{\sqrt{3}}\Rightarrow d\left( O,\left( SAC \right) \right)=\dfrac{2c}{\sqrt{3}}.\dfrac{\sqrt{30}}{20}=\dfrac{c}{\sqrt{10}}$
$\Rightarrow \dfrac{10}{{{c}^{2}}}=\dfrac{1}{{{h}^{2}}}+\dfrac{1}{{{c}^{2}}}\Rightarrow c=3h$
$\Rightarrow a+b+c=\dfrac{\sqrt{3}}{2}\Leftrightarrow h=\dfrac{\sqrt{3}}{12}\Rightarrow V=\dfrac{1}{3}.SO.{{S}_{\Delta ABC}}=\dfrac{1}{48}$
A. $\dfrac{1}{36}$.
B. $\dfrac{1}{48}$.
C. $\dfrac{1}{12}$.
D. $\dfrac{1}{24}$.
Đặt $d\left( A,BC \right)=a,d\left( B,AC \right)=b,d\left( C,AB \right)=c,SO=h$
Ta có ${{S}_{\Delta ABC}}={{S}_{\Delta OBC}}+{{S}_{\Delta OAC}}+{{S}_{\Delta OAB}}\Rightarrow a+b+c=\dfrac{\sqrt{3}}{2}\left( 1 \right)$
Mặt khác $\dfrac{d\left( O,\left( SBC \right) \right)}{d\left( A,\left( SBC \right) \right)}=\dfrac{OM}{AM}=\dfrac{OI}{AK}=\dfrac{2a}{\sqrt{3}}\Rightarrow d\left( O,\left( SBC \right) \right)=\dfrac{2a}{\sqrt{3}}.\dfrac{\sqrt{6}}{4}=\dfrac{a}{\sqrt{2}}$
$\Rightarrow \dfrac{2}{{{a}^{2}}}=\dfrac{1}{{{h}^{2}}}+\dfrac{1}{{{a}^{2}}}\Rightarrow a=h$
Tương tự $\dfrac{d\left( O,\left( SAC \right) \right)}{d\left( B,\left( SAC \right) \right)}=\dfrac{d\left( O,AC \right)}{d\left( B,AC \right)}==\dfrac{2b}{\sqrt{3}}\Rightarrow d\left( O,\left( SAC \right) \right)=\dfrac{2b}{\sqrt{3}}.\dfrac{\sqrt{15}}{10}=\dfrac{b}{\sqrt{5}}$
$\Rightarrow \dfrac{5}{{{b}^{2}}}=\dfrac{1}{{{h}^{2}}}+\dfrac{1}{{{b}^{2}}}\Rightarrow b=2h$
Tương tự $\dfrac{d\left( O,\left( SAB \right) \right)}{d\left( C,\left( SAB \right) \right)}=\dfrac{d\left( O,AB \right)}{d\left( C,AB \right)}==\dfrac{2c}{\sqrt{3}}\Rightarrow d\left( O,\left( SAC \right) \right)=\dfrac{2c}{\sqrt{3}}.\dfrac{\sqrt{30}}{20}=\dfrac{c}{\sqrt{10}}$
$\Rightarrow \dfrac{10}{{{c}^{2}}}=\dfrac{1}{{{h}^{2}}}+\dfrac{1}{{{c}^{2}}}\Rightarrow c=3h$
$\Rightarrow a+b+c=\dfrac{\sqrt{3}}{2}\Leftrightarrow h=\dfrac{\sqrt{3}}{12}\Rightarrow V=\dfrac{1}{3}.SO.{{S}_{\Delta ABC}}=\dfrac{1}{48}$
Đáp án B.