Gọi H là trung điểm của A B ⇒ S H ⊥ A B ⇒ S H ⊥ ( A B C D ) AB\Rightarrow SH\perp AB\Rightarrow SH\perp (ABCD) A B ⇒ S H ⊥ A B ⇒ S H ⊥ ( A B C D )
Suy ra HC là hình chiếu của SC lên ( A B C D ) ⇒ S C H ^ = 4 5 ∘ (ABCD)\Rightarrow \widehat{SCH}=45^{\circ} ( A B C D ) ⇒ S C H = 4 5 ∘
S A B C D = 2 a 2 S_{ABCD}=2a^{2} S A B C D = 2 a 2
S H = H C = 4 a 2 + a 2 4 = a 17 2 SH=HC=\sqrt{4a^{2}+\frac{a^{2}}{4}}=\frac{a\sqrt{17}}{2} S H = H C = 4 a 2 + 4 a 2 = 2 a 1 7
V S . A B C D = 1 3 . S H . S A B C D = 1 3 . a 17 2 . 2 a 2 = ∣ a 3 17 3 ∣ V_{S.ABCD}=\frac{1}{3}.SH.S_{ABCD}=\frac{1}{3}.\frac{a\sqrt{17}}{2}.2a^{2}=\left | \frac{a^{3}\sqrt{17}}{3} \right | V S . A B C D = 3 1 . S H . S A B C D = 3 1 . 2 a 1 7 . 2 a 2 = ∣ ∣ ∣ ∣ ∣ 3 a 3 1 7 ∣ ∣ ∣ ∣ ∣
d ( M , ( S A C ) ) = 1 2 d ( D , ( S A C ) ) = 1 2 d ( B , ( S A C ) ) = d ( H , ( S A C ) ) d(M,(SAC))=\frac{1}{2}d(D,(SAC))=\frac{1}{2}d(B,(SAC))=d(H,(SAC)) d ( M , ( S A C ) ) = 2 1 d ( D , ( S A C ) ) = 2 1 d ( B , ( S A C ) ) = d ( H , ( S A C ) )
Kẻ H I ⊥ A C , H K ⊥ S I ⇒ H K ⊥ A C ⇒ H K ⊥ ( S A C ) ⇒ d ( H ; ( S A C ) ) = H K . HI\perp AC,HK \perp SI\Rightarrow HK \perp AC\Rightarrow HK \perp (SAC)\Rightarrow d(H;(SAC))= HK. H I ⊥ A C , H K ⊥ S I ⇒ H K ⊥ A C ⇒ H K ⊥ ( S A C ) ⇒ d ( H ; ( S A C ) ) = H K .
Kẻ B E ⊥ A C ⇒ H I = 1 2 B E . 1 B E 2 = 1 B A 2 + 1 B C 2 = 1 a 2 + 1 4 a 2 = 5 4 a 2 ⇒ B E = 2 a 5 ⇒ H I = a 5 BE \perp AC\Rightarrow HI=\frac{1}{2}BE.\frac{1}{BE^{2}}=\frac{1}{BA^{2}}+\frac{1}{BC^{2}}=\frac{1}{a^{2}}+\frac{1}{4a^{2}}=\frac{5}{4a^{2}}\Rightarrow BE=\frac{2a}{\sqrt{5}}\Rightarrow HI=\frac{a}{\sqrt{5}} B E ⊥ A C ⇒ H I = 2 1 B E . B E 2 1 = B A 2 1 + B C 2 1 = a 2 1 + 4 a 2 1 = 4 a 2 5 ⇒ B E = 5 2 a ⇒ H I = 5 a
Từ đó suy ra: 1 H K 2 = 1 H I 2 + 1 H S 2 = 5 a 2 + 4 17 a 2 = 89 17 a 2 \frac{1}{HK^{2}}=\frac{1}{HI^{2}}+\frac{1}{HS^{2}}=\frac{5}{a^{2}}+\frac{4}{17a^{2}}=\frac{89}{17a^{2}} H K 2 1 = H I 2 1 + H S 2 1 = a 2 5 + 1 7 a 2 4 = 1 7 a 2 8 9
⇒ d ( M , ( S A C ) ) = a 17 89 = a 1513 89 \Rightarrow d(M,(SAC))=\frac{a\sqrt{17}}{\sqrt{89}}=\frac{a\sqrt{1513}}{89} ⇒ d ( M , ( S A C ) ) = 8 9 a 1 7 = 8 9 a 1 5 1 3
