Chương 1: KHỐI ĐA DIỆN

9.

\(\widehat{ABC}=60^0\Rightarrow\) các tam giác ABC và ACD đều 

Gọi M là trung điểm CD \(\Rightarrow AM\perp CD\Rightarrow CD\perp\left(SAM\right)\)

\(\Rightarrow\widehat{SMA}\) là góc giữa (SCD) và đáy

\(\Rightarrow\widehat{SMA}=60^0\)

\(AM=\dfrac{a\sqrt{3}}{2}\Rightarrow SA=AM.tan60^0=\dfrac{3a}{2}\)

\(V=\dfrac{1}{3}SA.2S_{ABC}=\dfrac{1}{3}\dfrac{3a}{2}.2.\dfrac{a^2\sqrt{3}}{4}=...\)

\(AB||CD\Rightarrow AB||\left(SCD\right)\Rightarrow d\left(AB;SC\right)=d\left(AB;\left(SCD\right)\right)=d\left(A;\left(SCD\right)\right)\)

Kẻ \(AH\perp SM\Rightarrow AH\perp\left(SCD\right)\Rightarrow AH=d\left(A;\left(SCD\right)\right)\)

\(\dfrac{1}{AH^2}=\dfrac{1}{SA^2}+\dfrac{1}{AM^2}=\dfrac{16}{9a^2}\Rightarrow AH=\dfrac{3a}{4}\)

\(\Rightarrow d\left(AB;SC\right)=\dfrac{3a}{4}\)

Hình vẽ bài 9:

10.

Gọi H là hình chiếu vuông góc của S lên (ABCD)

\(\Rightarrow H\in AB\)

Ta có \(AB^2=SA^2+SB^2\Rightarrow\Delta SAB\) vuông tại S

Áp dụng hệ thức lượng:

\(\dfrac{1}{SH^2}=\dfrac{1}{SA^2}+\dfrac{1}{SB^2}\Rightarrow SH=\dfrac{a\sqrt{3}}{2}\)

\(AM=\dfrac{1}{2}AB=a\Rightarrow S_{ADM}=\dfrac{1}{2}AD.AM=a^2\)

\(CN=\dfrac{1}{2}BC=a\Rightarrow S_{CDN}=\dfrac{1}{2}CN.CD=a^2\)

\(\Rightarrow S_{BMDN}=S_{ABCD}-\left(S_{ADM}+S_{CDN}\right)=4a^2-\left(a^2+a^2\right)=2a^2\)

\(\Rightarrow V_{S.BMDN}=\dfrac{1}{3}SH.S_{BMDN}=\dfrac{1}{3}.\dfrac{a\sqrt[]{3}}{2}.2a^2=\dfrac{a^3\sqrt{3}}{3}\)

\(SA\perp\left(ABCD\right)\Rightarrow\widehat{SBA}\) là góc giữa SB và đáy

\(\Rightarrow\widehat{SBA}=60^0\)

\(\Rightarrow SA=AB.tan60^0=a\sqrt{3}\)

\(\Rightarrow V_{SABCD}=\dfrac{1}{3}SA.AB.AD=\dfrac{2a^3\sqrt{3}}{3}\)

Do \(AB||CD\Rightarrow AB||\left(SCD\right)\Rightarrow d\left(B;\left(SCD\right)\right)=d\left(A;\left(SCD\right)\right)\)

Từ A kẻ \(AH\perp SD\Rightarrow AH\perp\left(SCD\right)\)

\(\Rightarrow AH=d\left(A;\left(SCD\right)\right)\)

\(\dfrac{1}{AH^2}=\dfrac{1}{SA^2}+\dfrac{1}{AD^2}=\dfrac{7}{12a^2}\Rightarrow AH=\dfrac{2a\sqrt{21}}{7}\)

\(\Rightarrow d\left(B;\left(SCD\right)\right)=\dfrac{2a\sqrt{21}}{7}\)

b.

\(BC||AD\Rightarrow MN||BC\)

\(\Rightarrow BCNM\) là hình thang

Lại có \(BC\perp\left(SAB\right)\Rightarrow BC\perp BM\)

\(\Rightarrow BCNM\) là hình thang vuông

Theo định lý Talet: \(\dfrac{MN}{AD}=\dfrac{SM}{SA}=\dfrac{2}{3}\Rightarrow MN=\dfrac{2}{3}AD=\dfrac{4a}{3}\)

\(BM=\sqrt{AB^2+AM^2}=\sqrt{AB^2+\left(\dfrac{SA}{3}\right)^2}=\dfrac{2a\sqrt{3}}{3}\)

\(\Rightarrow S_{BCNM}=\dfrac{1}{2}BM.\left(BC+MN\right)=\dfrac{10a^2\sqrt{3}}{9}\)

Từ S hạ \(SK\perp BM\Rightarrow SK\perp\left(BCMN\right)\)

\(SK=SM.cos\widehat{KSM}=\dfrac{2}{3}SA.\dfrac{AB}{BM}=a\)

\(\Rightarrow V_{SBCNM}=\dfrac{1}{3}SK.S_{BCNM}=\dfrac{10a^3\sqrt{3}}{27}\)

mọi người làm giúp mình với

\(SA\perp\left(ABCD\right)\Rightarrow\widehat{SMA}\) là góc giữa SM và đáy

\(\Rightarrow\widehat{SMA}=60^0\Rightarrow SA=AM.tan60^0=\sqrt{3a^2+\left(\dfrac{2a}{2}\right)^2}.\sqrt{3}=2a\sqrt{3}\)

Qua B kẻ đường thẳng song song AM cắt AD kéo dài tại E

\(\Rightarrow AM||\left(SBE\right)\Rightarrow d\left(AM;SB\right)=d\left(AM;\left(SBE\right)\right)=d\left(A;\left(SBE\right)\right)\)

Từ A kẻ \(AH\perp BE\) , từ A kẻ \(AK\perp SH\Rightarrow AK=d\left(A;\left(SBE\right)\right)\)

\(\widehat{DAM}=\widehat{AEB}\) (đồng vị) , mà \(\widehat{BAH}=\widehat{AEB}\) (cùng phụ \(\widehat{ABH}\))

\(\Rightarrow\widehat{DAM}=\widehat{BAH}\)

\(\Rightarrow AH=AB.cos\widehat{BAH}=AB.cos\widehat{DAM}=\dfrac{AB.AD}{AM}=\dfrac{2a.a\sqrt{3}}{2a}=a\sqrt{3}\)

\(\dfrac{1}{AK^2}=\dfrac{1}{AH^2}+\dfrac{1}{SA^2}=\dfrac{1}{3a^2}+\dfrac{1}{12a^2}=\dfrac{5}{12a^2}\)

\(\Rightarrow AK=\dfrac{2a\sqrt{15}}{5}\)

Áp dụng Pitago: \(\left\{{}\begin{matrix}SA^2+AB^2=SB^2=13a^2\\SA^2+AC^2=SC^2=20a^2\\AB^2+AC^2=BC^2=25a^2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}SA^2=4a^2\\AB^2=9a^2\\AC^2=16a^2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}SA=2a\\AB=3a\\AC=4a\end{matrix}\right.\)

\(V=\dfrac{1}{6}SA.AB.AC=4a^3\)

Lời giải:

Kẻ $SM\perp AB$. 

Mà $AB$ là giao tuyến của 2 mp vuông góc với nhau là $(SAB)$ và $(ABCD)$ nên $SM\perp (ABCD)$

$\Rightarrow \angle (SC, (ABCD))=\angle (SC, MC)=\widehat{SCM}$

Ta có:

$\frac{SM^2}{MC^2}=(\tan \widehat{SCM})^2=(\frac{\sqrt{15}}{5})^2=\frac{3}{5}$

$\Rightarrow 5SM^2=3MC^2$

Trong đó:

$SM^2=\frac{3}{4}AB^2$ do $SAB$ là tam giác đều

$MC^2=MB^2+BC^2=\frac{AB^2}{4}+a^2$

$\Rightarrow \frac{15}{4}AB^2=\frac{3}{4}AB^2+3a^2$

$\Rightarrow AB=a$ 

Vậy:

$SM^2=\frac{3}{4}AB^2=\frac{3}{4}a^2\Rightarrow SM=\frac{\sqrt{3}}{2}a$

$S_{ACD}=\frac{AD.AB}{2}=\frac{2a.a}{2}=a^2$

$V_{S.ABCD}=\frac{1}{3}.SM.S_{ABCD}=\frac{1}{3}.\frac{\sqrt{3}}{2}a.a^2=\frac{\sqrt{3}}{6}a^3$

Đáp án D.

Chọn D