It is known that the bottom of a pyramid p-abcd is a square. How many right triangles are there at most among the four sides? How many pairs of planes are perpendicular to each other

It is known that the bottom of a pyramid p-abcd is a square. How many right triangles are there at most among the four sides? How many pairs of planes are perpendicular to each other

As long as one side edge is perpendicular to the bottom, the four sides are all right triangles, which can be easily proved;
There are three pairs of sides that are perpendicular to each other

Square ABCD (side length is 4cm) and an isosceles right triangle EFG (bottom edge is 16cm) are in a straight line. Square ABCD moves along the straight line to the direction of the triangle at the speed of 4cm per second. When the square moves for 1 second, 2 seconds, 3 seconds... What is the overlapping area of the square and the triangle

If AB coincides with the bottom edge FG, and a and F are the same point, then when the square moves for 1 second, 2 seconds and 3 seconds, the overlapping area should be 16 square centimeters. Let FG be the bottom edge of the triangle, AB be on FG, the bottom edge is 16cm, and the square edge is 16cm

As shown in the figure, in the isosceles triangle ABCD, ab ∥ CD, AC = 10, CE ⊥ AB at point E, CD = 6cm, find the area of trapezoid ABCD

1. Isosceles trapezoid ABCD, CE vertical AB, and then DF vertical AB, then ADF equals BCE, take BCE away, get ADF left and become a rectangle, because AC equals 10, CE equals 6, so AE equals 8, so AE times CE is isosceles trapezoid, the area of abce equals 48

In the pyramid v-abcd, the bottom surface ABCD is a square with side length of 2, and the other four sides are isosceles triangles with side length of root 5
In the pyramid v-abcd, the bottom surface ABCD is a square with side length of 2, and the other four sides are isosceles triangles with side length of root 5, then the plane angle of the dihedral angle v-ab-c is

V as VE, AB as e, e as EF, CD as f
Company V, f
The angle VEF is the angle
VE=VF=EF=2
Angle VEF = 60 degrees

(2007 · Anhui) the square ABCD with side length of 2 is folded into a straight dihedral angle along the diagonal AC. after folding into a straight dihedral angle, the spherical distance between B and D is ()
A. 2πB. πC. π2D. π3

Let the midpoint of AC be o, then the distances from O to four points a, B, C, D are equal, ∧ o is the center of the ball, radius r = OA = 1, and ∠ BOD = π 2. The spherical distance between B and D is: π 2 × 1 = π 2. So choose C

Given an equilateral pyramid p-abcd, if its front view is an isosceles triangle whose side lengths are root 3, root 3 and root 2 respectively, calculate its surface area s and volume v

V=2*2*√2*(1/3)
=(4/3)√2
=1.89
S=2*2+2*√3*(1/2)*4
=4+4√3
=10.93

In the isosceles triangle ABCD, AB / / CD, diagonal AC vertical BD, CE is the height of trapezoid, if AB + CD = 6cm, then the area of trapezoid ABCD is

ABCD is isosceles trapezoid
∴AD=BC,∠DAB=∠CBA
∴△DAB≌△CBA
∴∠CAB=∠DBA
Let AC and BD intersect at point o,
∵AC⊥BD
The AOB is an isosceles right triangle
∴∠OAB=∠OBA=45°
CE is the height of trapezoid
The ∧ CEA is an isosceles right triangle
∵AB+CD=2AE=6cm
∴AE=3cm=CE
The area of trapezoidal ABCD = 6 * 3 / 2 = 9 (square centimeter)

The figure is on page 25 of edition a (fifth grade). The side length of square ABCD is 4cm. Calculate the area of rectangle efgd

(analysis: connect point a and point G to get the largest triangle AGD in the square, whose area is half of the square area and half of the rectangle area, then the area of square ABCD is equal to the area of rectangle efgd.)
That is: 4 * 4 = 16 (square centimeter) a: the area of rectangular efgd is 16 square centimeter

As shown in the figure, e is a point outside the square ABCD, and △ BCE is an equilateral triangle
Who knows,,

Connecting AC, AE
The quadrilateral ABCD is a square
∴AB=BC,∠ABC=90°
∵△ BCE is equilateral △ BCE
∴BC=BE,∠CBE=60°
∴BE=AB,∠ABE=150°
∴∠BAE=∠BEA=1/2（180°-150°）=15°
∵ AC is diagonal
∴∠BAC=45°
∴∠CAE=30°

If square e is a point in square ABCD and △ BCE is an equilateral triangle, then ∠ Abe =, ∠ AEB=

Δ BCE is equilateral, so ∠ ECB = 60 °
And ∠ ABC = 90 degree, so ∠ Abe = ∠ ABC - ∠ ECB = 30 degree
A square ABCD has ab = BC
Δ BCE is an equilateral triangle with BC = be
So AB = be
In the isosceles triangle Abe, ∠ AEB = (180 ° - ABE) / 2 = 150 ° / 2 = 75 °