Question

求大神解答几何18题

Answer 1

易知四边形A1ABB1是菱形
所以,OS⊥A1B
BC⊥平面A1ABB1,A1B是斜线A1C射影
所以AB1⊥A1C
因AB1⊥BC,A1B和A1C交于点A1
所以,AB1⊥平面A1BC
2)连接AC1
AC=5,BC=3
所以,AB=A1A=A1B=4
A1C^2=A1B^2+BC^2=4^2+3^2=25
A1C=5,AC=5
A1C=AC=5
三角形AA1C是等腰三角形,高=√(AC^2-1/4*A1A^2)=√(25-4)=√21

Saa1c1=Saa1c=1/2*√21*4=2√21
Saa1b1=1/2*4*4*√3/2=4√3
Vb1--aa1c1=Vc1--aa1b1

1/3*Saa1c1*H=1/3*Saa1b1*B1C1
H=Saa1b1*B1C1/Saa1c1=3*4√3/2√21=6√7/7
H=6√7/7
点B1到平面A1ACC1的距离为:6√7/7

Answer 2

等下