13.
\(BH=\dfrac{1}{3}AB=\dfrac{a}{3}\)
Áp dụng định lý hàm cos trong tam giác BCH:
\(CH=\sqrt{BH^2+BC^2-2BH.BC.cos60^0}=\dfrac{a\sqrt{7}}{3}\)
\(SH\perp\left(ABC\right)\Rightarrow\widehat{SCH}\) là góc giữa SC và (ABC)
\(\Rightarrow\widehat{SCH}=60^0\)
\(\Rightarrow SH=CH.tan\widehat{SCH}=\dfrac{a\sqrt{21}}{3}\)
\(\Rightarrow V=\dfrac{1}{3}SH.\dfrac{a^2\sqrt{3}}{4}=\dfrac{a^3\sqrt{7}}{12}\)
13b.
Qua A kẻ đường thẳng d song song BC
Từ H hạ \(HD\perp d\)
Ta có \(BC||AD\Rightarrow BC||\left(SAD\right)\)
\(\Rightarrow d\left(BC;SA\right)=d\left(BC;\left(SAD\right)\right)=d\left(B;\left(SAD\right)\right)\)
Mà \(\left\{{}\begin{matrix}BH\cap\left(SAD\right)=A\\AH=\dfrac{2}{3}AB\end{matrix}\right.\)
\(\Rightarrow d\left(H;\left(SAD\right)\right)=\dfrac{2}{3}d\left(B;\left(SAD\right)\right)\Rightarrow d\left(B;\left(SAD\right)\right)=\dfrac{3}{2}d\left(H;\left(SAD\right)\right)\)
Từ H kẻ \(HK\perp SD\) \(\Rightarrow HK\left(SAD\right)\)
\(\Rightarrow HK=d\left(H;\left(SAD\right)\right)\)
\(\widehat{DAH}=\widehat{B}=60^0\) (so le trong) \(\Rightarrow DH=AH.sin60^0=\dfrac{a\sqrt{3}}{3}\)
\(\dfrac{1}{HK^2}=\dfrac{1}{SH^2}+\dfrac{1}{DH^2}=\dfrac{22}{7a^2}\)
\(\Rightarrow HK=\dfrac{a\sqrt{154}}{22}\Rightarrow d\left(SA;BC\right)=\dfrac{3}{2}HK=...\)
14.
Áp dụng định lý hàm cos:
\(cos\widehat{SCA}=\dfrac{SC^2+AC^2-SA^2}{2SC.AC}=\dfrac{1}{4}\)
\(\Rightarrow CH=AC.cos\widehat{SCA}=\dfrac{a}{4}=\dfrac{1}{8}SC\)
\(\Rightarrow d\left(H;\left(ABC\right)\right)=\dfrac{1}{8}d\left(S;\left(ABC\right)\right)\)
\(\Rightarrow V_{H.ABC}=\dfrac{1}{8}V_{SABC}\)
\(\Rightarrow V_{S.ABH}=\dfrac{7}{8}V_{SABC}\)
Gọi O là tâm đáy \(\Rightarrow OA=\dfrac{2}{3}.\dfrac{a\sqrt{3}}{2}=\dfrac{a\sqrt{3}}{3}\)
\(\Rightarrow SO=\sqrt{SA^2-OA^2}=\dfrac{a\sqrt{33}}{3}\)
\(V_{SABC}=\dfrac{1}{3}SO.\dfrac{a^2\sqrt{3}}{4}=\dfrac{a^3\sqrt{11}}{12}\)
\(\Rightarrow V_{S.ABH}=\dfrac{7a^3\sqrt{11}}{96}\)
