a.

Gọi H là trung điểm AB \(\Rightarrow SH\perp AB\Rightarrow SH\perp\left(ABCD\right)\) \(\Rightarrow SH\perp CD\)

Gọi E là trung điểm CD \(\Rightarrow HE||BC\Rightarrow HE\perp CD\)

\(\Rightarrow CD\perp\left(SHE\right)\)

Từ H kẻ \(HF\perp SE\)

\(\Rightarrow HF\perp\left(SCD\right)\Rightarrow HF=d\left(H;\left(SCD\right)\right)\)

\(SH=\dfrac{a\sqrt{3}}{2}\) (trung tuyến tam giác đều cạnh a), \(HE=BC=a\)

Hệ thức lượng: \(HF=\dfrac{SH.HE}{\sqrt{SH^2+HE^2}}=\dfrac{a\sqrt{21}}{7}\)

Do \(AH||CD\Rightarrow AH||\left(SCD\right)\Rightarrow d\left(A;\left(SCD\right)\right)=d\left(H;\left(SCD\right)\right)=\dfrac{a\sqrt{21}}{7}\)

b.

Theo tính chất trọng tâm, ta có \(GS=\dfrac{2}{3}HS\)

Mà \(HG\cap\left(SCD\right)=S\Rightarrow d\left(G;\left(SCD\right)\right)=\dfrac{2}{3}d\left(H;\left(SCD\right)\right)=\dfrac{2a\sqrt{21}}{21}\)

c.

Từ H kẻ \(HK\perp SA\)

Do \(\left\{{}\begin{matrix}SH\perp\left(ABCD\right)\Rightarrow SH\perp AD\\AD\perp AB\end{matrix}\right.\) \(\Rightarrow AD\perp\left(SAB\right)\)

\(\Rightarrow AD\perp HK\)

\(\Rightarrow HK\perp\left(SAD\right)\Rightarrow HK=d\left(H;\left(SAD\right)\right)\)

Hệ thức lượng: \(HK=\dfrac{SH.AH}{\sqrt{SH^2+AH^2}}=\dfrac{a\sqrt{3}}{4}\)

Do \(BC||AD\Rightarrow BC||\left(SAD\right)\Rightarrow d\left(BC;SD\right)=d\left(BC;\left(SAD\right)\right)=d\left(B;\left(SAD\right)\right)\)

Mà \(\left\{{}\begin{matrix}BH\cap\left(SAD\right)=A\\BA=2HA\end{matrix}\right.\)

\(\Rightarrow d\left(BC;SD\right)=d\left(B;\left(SAD\right)\right)=2d\left(H;\left(SAD\right)\right)=\dfrac{a\sqrt{3}}{2}\)

d.

Gọi M là trung điểm AD \(\Rightarrow OM||CD\Rightarrow CD||\left(SOM\right)\)

\(\Rightarrow d\left(CD;SO\right)=d\left(CD;\left(SOM\right)\right)=d\left(E;\left(SOM\right)\right)\)

Mà \(\left\{{}\begin{matrix}HE\cap\left(SOM\right)=O\\HO=EO\end{matrix}\right.\)

\(\Rightarrow d\left(E;\left(SOM\right)\right)=d\left(H;\left(SOM\right)\right)\)

Từ H kẻ \(HI\perp SO\)

\(OM||CD\Rightarrow OM\perp\left(SHE\right)\Rightarrow OM\perp HI\)

\(\Rightarrow HI\perp\left(SOM\right)\Rightarrow HI=d\left(H;\left(SOM\right)\right)\)

\(OH=\dfrac{1}{2}BC=\dfrac{a}{2}\)

Hệ thức lượng:

\(HI=\dfrac{SH.HO}{\sqrt{SH^2+HO^2}}=\dfrac{a\sqrt{3}}{4}\)

Chọn: B

a: BC vuông góc SA
BC vuông góc AB

=>CB vuông góc (SBA)

DC vuông góc AD

DC vuông góc SA

=>DC vuông góc (SAD)

=>(SDC) vuông góc (SAD)

b: (SC;(SAD))=(SC;SD)=góc CSD

\(SD=\sqrt{SA^2+AD^2}=2a\sqrt{7}\)

\(AC=\sqrt{\left(2a\right)^2+3a^2}=a\sqrt{7}\)

\(SC=\sqrt{SA^2+AC^2}=4a\sqrt{2}\)

\(cosCSD=\dfrac{SC^2+SD^2-DC^2}{2\cdot SC\cdot SD}=\dfrac{32a^2+28a^2-4a^2}{2\cdot2a\sqrt{7}\cdot4a\sqrt{2}}=\dfrac{\sqrt{14}}{4}\)

=>góc CSD=21 độ

(SC;(ABCD))=(CS;CA)=góc SCA

tan SCA=SA/AC=5/căn 7

=>góc SCA=62 độ

Đáp án D

Đáp án D

Hạ 

Hạ