As shown in the figure, pdce is rectangular, ABCD is trapezoidal, plane pdce ⊥ plane ABCD, ∠ bad = ∠ ADC = 90 # 186;, ab = ad = # 189; CD = a, PD = √ 2 (1) If M is the midpoint of PA, the MDE of AC ‖ plane is proved; (2) Find the size of the sharp dihedral angle between planar pad and PBC

As shown in the figure, pdce is rectangular, ABCD is trapezoidal, plane pdce ⊥ plane ABCD, ∠ bad = ∠ ADC = 90 # 186;, ab = ad = # 189; CD = a, PD = √ 2 (1) If M is the midpoint of PA, the MDE of AC ‖ plane is proved; (2) Find the size of the sharp dihedral angle between planar pad and PBC

Connect PC, connect De to N, connect Mn,
In △ PAC, m and N are the midpoint of PA and PC
MN∥AC
Because Mn &; MDE,
AC ‖ plane mde
Take D as the origin of the space coordinate system, and take the straight lines of Da, DC and DP as the X, y and Z axes to establish the space rectangular coordinate system,
Then p (0,0, radical 2a), B (a, a, 0), C (0,2a, 0), and,
Because the phasor cannot be printed
Pb vector = (a, a, radical 2a), BC vector = (- A, a, 0),
The unit normal vector of planar pad is m = (0,1,0), and the normal vector of planar PBC is n = (x, y, 1)
N vector * Pb vector = ax + ay - radical 2A = 0
N vector * BC vector = - ax + ay = 0
X = y = radical 2 / 2
N vector = (radical 2 / 2, radical 2 / 2,1)
Let the sharp dihedral angle between planar pad and planar PBC be θ,
Cos θ = (M vector * n vector) / im vector I * in vector I = 1 / 2
The sharp dihedral angle of planar pad and PBC is cos θ = 1 / 2

As shown in the figure, plane pad ⊥ plane ABCD, ABCD is a square, PA ⊥ ad, and PA = ad = 2, e, F, G are the midpoint of line PA, PD, CD respectively. (1) prove: BC ∥ plane EFG; (2) calculate the volume of triangular pyramid e-afg

(1) It is proved that: ∵ e, f are the midpoint of line PA, PD respectively, ∵ EF ∥ ad (2 points) ABCD is a square, BC ad, BC EF (4 points) and ∵ BC ⊄ plane EFG, EF ⊂ plane EFG, ∵ BC ∥ plane EFG & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & n

As shown in the figure, the planar pad is perpendicular to the planar ABCD, the quadrilateral ABCD is a square, the triangular pad is a right triangle, and PA = ad = 2, e, F, G are line segments PA respectively
It's about the college entrance examination

The complete title is found
The planar pad is perpendicular to the planar ABCD, the quadrilateral ABCD is a square, the triangular pad is a right triangle, and PA = ad = 2, e, F and G are the midpoint of the segments PA, PD and CD respectively. It is proved that Pb is parallel to the planar EFG, and the cosine of the angle between the straight line eg and BD on the different plane is obtained
Take the midpoint of AB as h, connect EH and GH, in △ PAB, eh ‖ Pb, eh in efgh, Pb out of efgh, EFG in ∥ Pb ‖ plane. Connect AG with BD in M, make Mn ‖ eg in ⊿ age and PA in N, then ∠ DMN is obtained. According to the cosine theorem, the cosine value of the angle between eg and BD is √ 3 / 6

Point P is outside the plane of square ABCD, triangle PDC and triangle pad are two congruent isosceles triangles, and PD is perpendicular to ad, PD is perpendicular to ad, then PA
Point P is outside the plane of square ABCD, triangle PDC and triangle pad are two congruent isosceles triangles, and PD is perpendicular to DC, PD is perpendicular to ad, then the degree of angle between PA and BD is () please explain the reason

Let the intersection of AC and BD be o, take the midpoint e of PC, connect de and OE, then
According to the meaning of the title, RT △ PDC ≌ RT △ pad,
So PA = PC = √ 2 * ad = BD,
Because PE = EC, de = PC / 2,
Because Ao = OC, so OE ‖ PA, OE = PA / 2,
Because do = BD / 2, PA = PC = = BD,
So de = od = OE, △ deo is an equilateral triangle,
So the angle between PA and BD = the angle between OE and BD = 60 °

In the pyramid p-abcd, the bottom surface ABCD is a rectangle, the side edge PA is perpendicular to the bottom surface, and E and F are the midpoint of AB and PC respectively
In the pyramid p-abcd, the bottom surface ABCD is a rectangle, the side edge PA is perpendicular to the bottom surface, and E and F are the midpoint of AB and PC respectively. (1) calculate EF / / planar pad; (2) when the dihedral angle between planar PCD and planar ABCD is large, the straight line EF is perpendicular to planar PCD?

Take the midpoint g of PD, connect FG and Ag, then
FG//1/2CD,AE//1/2CD
Ψ FG / / AE and FG = AE
The quadrilateral aefg is a parallelogram
∴EF//AG
Ψ EF / / planar pad
(2)
The bottom ABCD is rectangular
∴CD⊥AD
∵ the side edges are perpendicular to the bottom
⊥ CD ⊥ pad
∴CD⊥PD
The plane PCD and the plane ABCD form a dihedral angle
∵ EF ⊥ PCD
∴EF⊥CD
Also EF / / Ag
⊥ Ag ⊥ PCD
⊥ Ag ⊥ PD and point G is the midpoint of PD, PA ⊥ ad
∴∠PDA=45°
So the dihedral angle between PCD and ABCD is 45 degrees

Four pyramid p-abcd, AB vertical ad, CD vertical ad, PA vertical bottom ABCD, PA equal to ad equal to 2Ab equal to 2, M is the midpoint of PC, find the BM parallel plane pad

The condition is not complete. It should be pa = ad = CD = 2Ab = 2
Take PD midpoint n
Then Mn is the median of △ PCD
MN∥CD
MN=1/2CD = 1
∵BA⊥AD,CD⊥AD
∴BA∥CD
∴MN∥AB,MN=AB
The quadrilateral mnab is a parallelogram
BM∥AN
It is also in the pad
｜ BM ‖ pad

As shown in the figure, it is known that the bottom surface ABCD of the pyramid p-abcd is a parallelogram, m and N are the midpoint of edges AB and PC respectively, and the intersection of plane CMN and plane pad with PE is proved: (1) Mn ∥ plane pad; (2) Mn ∥ PE

It is proved that: (1) as shown in the figure, take the midpoint Q of DC, connect MQ, NQ. ∵ n, q is the midpoint of PC and DC respectively, ∵ NQ ⊄ planar pad, PD ⊂ planar pad, ∵ NQ ⊂ planar pad, ∵ m is the midpoint of AB, quadrilateral ABCD is parallelogram, ∵ MQ ⊄ planar pad, ad ⊂ planar pad

As shown in the figure, in the pyramid p-abcd, ab ∥ CD, CD = 2Ab, ab ⊥ plane pad, e is the midpoint of PC. (1) prove: be ∥ plane pad; (2) if ad ⊥ Pb, prove: PA ⊥ plane ABCD

It is proved that: (1) take the midpoint F of PD and connect EF and AF. because e is the midpoint of PC and F is the midpoint of PD, EF ‖ CD, and CD = 2ef. And because ab ‖ CD, CD = 2Ab, EF ‖. AB, that is, the quadrilateral abef is a parallelogram. So be ‖ AF (5) AF ⊂ planar pad, be ⊄ planar pad

In a right triangle ABC, if the angle c is equal to 90 degrees, the perimeter is 60, and the ratio of the hypotenuse to a right side is 13:5, then the lengths of the three sides of the triangle are?

The ratio of hypotenuse to a right angle is 13:5, 13 & sup2; - 5 & sup2; = 12 & sup2; then the second right angle should be 12, Pythagorean theorem
12+13+5=30 60/30=2
12*2=24 13*2=26 5*2=10

As shown in the figure, at the intersection of triangular pyramid s-abc, plane efghbc, CA, as, Sb and points e, F, G, h, and SA ⊥ plane efgh, SA ⊥ AB, EF ⊥ FG. (1) ab ⊥ plane efgh; (2) GH ⊥ EF; (3) GH ⊥ plane sac

(1) ⊥ SA ⊆ plane efgh, GH ⊆ plane efgh, ⊆ SA ⊥ GH ⊙ in plane SAB, SA ⊥ AB, ⊈ ab ⊈ plane efgh, GH ⊆ plane efgh, ⊆ ab ⊉ plane efgh (6 points) (2) ∵ ab ∥ plane efgh, ab ⊆ plane ABC, plane ABC ∩ plane efgh = EF ∩ ab ∥ EF ∥ ab ∥ GH, ∩ EF ∥ GH (10 points) (3) ∵ SA ⊆ plane efgh, SA ⊆ plane sac ⊥ plane efgh, the intersection line is FG ∵ EF ⊥ GH, EF ⊥ FG, ⊆ GH ⊆ plane efgh, ⊥ GH ⊥ plane sac (140 points)