+ Tính thể tích khối chóp:
Ta có: S A B C D = a 2 S_{ABCD}=a^{2} S A B C D = a 2
( S B , ( A B C D ) ) = ( S B ; B A ) = S B A ^ = 6 0 ∘ (SB,(ABCD))=(SB;BA)=\widehat{SBA}=60^{\circ} ( S B , ( A B C D ) ) = ( S B ; B A ) = S B A = 6 0 ∘
tan 6 0 ∘ = S A B A ⇒ S A = B A . tan 6 0 ∘ = a 3 \tan 60^{\circ}=\frac{SA}{BA}\Rightarrow SA=BA.\tan 60^{\circ}=a\sqrt{3} tan 6 0 ∘ = B A S A ⇒ S A = B A . tan 6 0 ∘ = a 3
Thể tích V S . A B C D = 1 3 S A . S A B C D = 3 a 3 3 V_{S.ABCD}=\frac{1}{3}SA.S_{ABCD}=\frac{\sqrt{3}a^{3}}{3} V S . A B C D = 3 1 S A . S A B C D = 3 3 a 3
+ Tính khoảng cách từ trọng tâm G của tam giác SAD đến (SBD)
Gọi O = A C ∩ B D , O=AC\cap BD, O = A C ∩ B D , { B D ⊥ A C B D ⊥ S A ⇒ B D ⊥ ( S A C ) \left\{\begin{matrix} BD\perp AC\\BD\perp SA \end{matrix}\right.\Rightarrow BD\perp (SAC) { B D ⊥ A C B D ⊥ S A ⇒ B D ⊥ ( S A C )
Kẻ A H ⊥ S O AH\perp SO A H ⊥ S O { A H ⊥ S O A H ⊥ B D ⇒ A H ⊥ ( S B D ) \left\{\begin{matrix} AH\perp SO\\ AH\perp BD \end{matrix}\right.\Rightarrow AH\perp (SBD) { A H ⊥ S O A H ⊥ B D ⇒ A H ⊥ ( S B D )
d ( A , ( S B D ) ) = A H d(A,(SBD))=AH d ( A , ( S B D ) ) = A H
Kẻ G K ⊥ H M , GK\perp HM, G K ⊥ H M , G K / / A H ⇒ G K ⊥ ( S B D ) GK//AH\Rightarrow GK\perp (SBD) G K / / A H ⇒ G K ⊥ ( S B D )
d ( M , ( S B D ) ) = G K d(M,(SBD))=GK d ( M , ( S B D ) ) = G K
Gọi M là trung điểm SD ta có d ( G ; ( S B D ) ) d ( A ; ( S B D ) ) = G K A H = M G M A = 1 3 \frac{d(G;(SBD))}{d(A;(SBD))}=\frac{GK}{AH}=\frac{MG}{MA}=\frac{1}{3} d ( A ; ( S B D ) ) d ( G ; ( S B D ) ) = A H G K = M A M G = 3 1
Ta có d ( M , ( S B D ) ) = G K = 1 3 A H = 1 3 1 1 S A 2 + 1 A O 2 = 1 3 1 1 3 a 2 + 1 a 2 = a 21 21 d(M,(SBD))=GK=\frac{1}{3}AH=\frac{1}{3}\sqrt{\frac{1}{\frac{1}{SA^{2}}+\frac{1}{AO^{2}}}}=\frac{1}{3}\sqrt{\frac{1}{\frac{1}{3a^{2}}+\frac{1}{a^{2}}}}=\frac{a\sqrt{21}}{21} d ( M , ( S B D ) ) = G K = 3 1 A H = 3 1 S A 2 1 + A O 2 1 1 = 3 1 3 a 2 1 + a 2 1 1 = 2 1 a 2 1
