Chương 1: KHỐI ĐA DIỆN

\(V=\dfrac{1}{3}SA.a^2=\dfrac{\sqrt{2}}{3}a^3\)

\(\Rightarrow SA=a\sqrt{2}\)

\(\Rightarrow tan\widehat{SCA}=\dfrac{SA}{AC}=1\)

\(\Rightarrow\widehat{SCA}=45^0\)

c

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) \(\Rightarrow DH=AH.sin60^0=\dfrac{a\sqrt{3}}{3}\)

\(\dfrac{1}{HK^2}=\dfrac{1}{SH^2}+\dfrac{1}{DH^2}=\dfrac{22}{7a^2}\)

\(\Rightarrow HK=\dfrac{a\sqrt{154}}{22}\Rightarrow d\left(SA;BC\right)=\dfrac{3}{2}HK=...\)

Hình vẽ bài 13:

11.

Áp dụng định lý hàm cos:

\(BC=\sqrt{AB^2+AC^2-2AB.AC.cos120^0}=a\sqrt{7}\)

Áp dụng công thức trung tuyến:

\(AM=\dfrac{\sqrt{2\left(AB^2+AC^2\right)-BC^2}}{2}=\dfrac{a\sqrt{3}}{2}\)

Do \(SM\perp\left(ABC\right)\Rightarrow\widehat{SAM}=45^0\)

\(\Rightarrow SM=AM.tan45^0=\dfrac{a\sqrt{3}}{2}\)

\(V=\dfrac{1}{3}SM.\dfrac{1}{2}AB.AC.sin120^0=\dfrac{a^3}{4}\)

Hình vẽ bài 11:

12.

Do \(AA'||CC'\Rightarrow\) góc giữa AA' và BC' bằng góc giữa BC' và CC'

\(\Rightarrow\widehat{BC'C}=30^0\)

Gọi D là trung điểm BC \(\Rightarrow AD\perp BC\Rightarrow AD\perp\left(BCC'B'\right)\)

\(\Rightarrow AD=d\left(A;\left(BCC'B'\right)\right)\)

\(AA'||\left(BCC'B'\right)\Rightarrow d\left(AA';BC'\right)=d\left(AA';\left(BCC'B'\right)\right)=d\left(A;\left(BCC'B'\right)\right)=AD\)

\(\Rightarrow AD=a\)

Mà \(AD=\dfrac{BC\sqrt{3}}{2}\Rightarrow BC=\dfrac{2a\sqrt{3}}{3}\)

\(\Rightarrow CC'=\dfrac{BC}{tan30^0}=2a\)

\(\Rightarrow V=CC'.\dfrac{BC^2\sqrt{3}}{4}=\dfrac{2a^3\sqrt{3}}{3}\)