The square ABCD is folded into a dihedral angle with a plane angle of 120 ° along the diagonal BD to calculate the size of the dihedral angle b-ac-d The square ABCD is folded into a dihedral angle with a plane angle of 120 ° along the diagonal BD Finding the size of dihedral angle b-ac-d

The square ABCD is folded into a dihedral angle with a plane angle of 120 ° along the diagonal BD to calculate the size of the dihedral angle b-ac-d The square ABCD is folded into a dihedral angle with a plane angle of 120 ° along the diagonal BD Finding the size of dihedral angle b-ac-d

Let the midpoint of BD be o, the midpoint of AC in geometry be p, ab = 1, connect Ao, Co, BP and DP. Easy proof: Ao ⊥ BD, Co ⊥ BD ∠ AOC is the plane angle of a-bd-c ∠ AOC = 120 ° AC = 2 * (√ 2 / 2cos60 °) = √ 6 / 2. Easy proof: AC ⊥ BP, AC ⊥ DP ∠ BPD are dihedral angles. Plane angle of b-ac-d BP = DP = √ (1 - (√ 6 / 4) ^ 2) = √

How to make plane angle of dihedral angle

This is the definition. If you cross any point of the intersection line of two half planes, you can make a vertical line of the intersection line in the two planes. Sometimes you can find it according to the theorem of three vertical lines

If the two half planes of a dihedral angle are respectively perpendicular to the two half planes of another dihedral angle, the relationship between the two dihedral angles is ()
A. Equal B. equal or complementary C. complementary D. indeterminate

If the half planes of two dihedral angles correspond to perpendicularity respectively, then the two dihedral angles are equal or complementary "(properties of faces and dihedral angles), but this proposition is not necessarily correct, as shown in the following figure is a counter example: in the cube abcd-a1b1c1d1, the two half planes of dihedral angle d-aa1-f and dihedral angle d1-dc-a correspond to perpendicularity respectively, but the two dihedral angles are not equal, So choose: D

If there is a line in one half plane of a dihedral angle of 45 ° and the edge of the dihedral angle forms a 45 ° angle, then the line and the other half plane of the dihedral angle form a 45 ° angle

Included angle = 30 degree

In rectangular trapezoid ABCD, ab ∥ CD, ad ⊥ AB, ∠ B = 45 °, ab = 2CD = 2, M is the midpoint of waist BC, then Ma · MD=______ ．

Take a as the origin, AB as the x-axis, ad as the y-axis, establish the rectangular coordinate system. Then: a (0,0), B (2,0), D (0,1), C (1,1), m (32,12). Because AB = 2CD = 2, ∠ B = 45, ad = DC = 1, M is the midpoint of waist BC, then the distance between m point and ad = 12 (DC + AB) = 32, the distance between m point and ab = 12 (DC + AB) = 32, the distance between m point and ab = 1, the distance between m point and ab = 12 (DC + AB) = 32

The quadrilateral ABCD is a right angle trapezoid, the waist AB perpendicular to the bottom is a circle with diameter tangent to the waist CD, the length of AB is 8, the perimeter of trapezoid ABCD is 28, and the upper bottom AD and the lower bottom BC are calculated
As shown in the figure, the quadrilateral ABCD is a right angled trapezoid, and the circle with the diameter of the waist AB perpendicular to the bottom is tangent to the waist CD. It is known that the length of AB is 8cm, and the perimeter of the trapezoid ABCD is 28cm. Find the length of the upper bottom ad and the lower bottom BC of the trapezoid

Answer: the upper bottom is 2 cm, and the lower bottom is 8 cm
Because the quadrilateral is a right angle trapezoid, and ab is the diameter. Take the midpoint e of AB side to connect CE, and De, then the angle CED is a right angle, so be * be + BC * BC = EC * EC (1) AE * AE + ad * ad = de * de (2)
Because de * de + EC * EC = CD * CD (3)
So the sum on the left side of (1) (2) is equal to the sum on the right side of (3), and because the perimeter is 28 and ab is 8, there is AD + CD + BC = 20 to make the vertical F of CD through e, ad = DF and BC = FC
So CD = 10
Set of equations: AD + BC = 10
(AD*AD+AE*AE)+(BE*BE+BC*BC)=ED*ED+EC*EC=CD*CD=100
The solution is ad = 2, BC = 8

Isosceles trapezoid ABCD, ad parallel CB, ∠ B = 60 °, ab = CD, ad = 15, BC = 29, then how long is its waist

The straight line AE ⊥ BC is made through point a, and the straight line DF ⊥ BC is made through point D
∵AB∥BC AE⊥BC DF⊥BC
∴AD=EF=15
And ∵ AB = CD ∠ B = 60
∴∠BAE=30 ° AE=FC=1/2(BC-EF)=7
∴AB=CD=2BE=2FC=2X7=14

As shown in the figure, in the trapezoidal ABCD, ad ‖ CB, ∠ C = 90 ° and ab + BC = AB, AB is the diameter of circle o
There is no trapezoidal median line in our textbook

Extend BC to intercept CF = ad, connect AF to CD and E ∵ ad ∥ BC ∥ DAE = ∥ F, ∥ d = ∥ ECF = 90 °≌ ade ≌ FCE ∥ AE = EF, connect be ∵ AD + BC = AB, ∥ BF = BC + CF = ab ∥ ABF is isosceles triangle ∥ be ⊥ AF, connect OE ∥ in RT △ Abe, O is the midpoint of AB, then OE = 1 / 2Ab = OA = ob ∥ OE is diameter

As shown in the figure, in the right angle trapezoid ABCD, ad ∥ BC, ∠ C = 90 ° and ab > AD + BC, AB is the diameter of ⊙ o, then the position relationship between the straight line CD and ⊙ o is ()
A. Separation B. tangency C. intersection D. indeterminate

Make OE ⊥ CD in E. ∵ ad ∥ BC, ∠ C = 90 °, OE ⊥ CD, ∥ ad ∥ OE ∥ BC, OA = ob, ∥ de = CE, ∥ OE = AD + BC2, AB > AD + BC, ∥ OE < AB2, that is, the distance from the center of a circle to the straight line is less than the radius of the circle, then the straight line intersects the circle

As shown in the figure, the center of circle O is on the bottom edge ab of ladder ABCD and is tangent to the other three sides. If AB = 10 and ad = 6, then CB length ()
A. 4b. 5C. 6D

Let the radius of circle o be r, and the circle O is tangent to ad, DC and CB at points e, F and h, connecting OE, OD, of, OC and oh. Let CD = y, CB = X. let s trapezoid ABCD = s, then s = 12 (CD + AB) r = 12 (y + 10) r --- - (1) s = s △ BOC + s △ cod + s △ DOA = 12xr + 12yr + 12 × 6R --- - (2) simultaneous (1) (2) is obtained