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Given that points a and B are on the same side of plane α and the distances to plane α are D and 3D respectively, then the distance from the midpoint of a and B to plane α is______ ．
Make ad ⊥ α in D, BC ⊥ α in C, connect DC, then take the midpoint m and N of AB and DC respectively, connect Mn ⊥ ad ⊥ α, BC ⊥ α ⊥ ad ∥ BC, Mn is trapezoidal, the median line of ABCD ⊥ Mn = 12 (AD + BC) ⊥ ad ⊥ α, BC ⊥ α ∥ nm ∥ BC ∥ ad, Mn is the distance from the midpoint of AB to plane α, and AD and BC are points a and B respectively
If line AB is on the same side of plane α and the distances from a and B to α are 3 and 5 respectively, then the distance from the midpoint of AB to α is 0______ ．
Let AC ⊥ plane α, BD ⊥ plane α, then acdb ⊥ plane α, and PE ⊥ CD through P, then PE represents the distance from P to α. From the plane geometry knowledge, we can see that PE is the median line of trapezoid, so PE = 3 + 52 = 4, so the answer is: 4
Can dihedral angle be solved by geometric method without space vector
1. Find out the dihedral angle with the angle of shadow skew, and then calculate it;
2. Finding dihedral angle by congruent triangle method;
3. Using the formula of projective area;
Find out dihedral angle by vertical plane method of edge
5. Use the definition of dihedral angle;
The angle between a plane and a plane in solid geometry
In the cuboid abcd-a1b1c1d1, ab = 4, ad = 2, Aa1 = 3, find the angle between AC1 and plane ABCD, plane add1a1, plane abb1a1
The answer is 48 degrees, 22 degrees
(1) The angle between AC1 and plane ABCD: namely ∠ CaC1, AC ⊥ C1C,
AC & # 178; = AB & # 178; + BC & # 178; = 16 + 4 = 20, Tan ∠ CaC1 = C1C / AC solvable ∠ CaC1
(2) Angle with plane add1a1: namely ∠ d1ac1, (3) angle with plane abb1a1: namely ∠ b1ac1
Due to the word limit, the understanding is not clear
AB is the diameter of circle O, the intersection of chord CD and extension line of AB is point P, and DP = ob, the degree of arc AC is 84 ° and the degree of ∠ P is calculated
Let P = x, DOP = P = X
OCD=ODC=2x
AOC=3x=84 x=28
In circle O, the chords AC and BD intersect at point E, and arc AB = arc BC = arc CD,
Arc AB = arc BC, arc BC = arc CD
∴AB=BC=CD
∴∠BAC=∠BCA=∠CBD=∠CDB
∵∠BEC=130°
∴∠BCA=∠CBD=25°,∠CED=50°
∴∠ACD=180°-50°-25°=105°．
As shown in the figure, AB is the diameter of O, ab = 4, OC is the radius of O, OC ⊥ AB, point D is on arc AC, arc ad = 2, arc CD, if point P is a moving point on radius OC, then
Then the minimum value of AP + PD is
Let OP = x, then 0 ≤ x ≤ 2 by Pythagorean theorem: AP ^ 2 = OA ^ 2 + OP ^ 2 = 4 + x ^ 2, because arc ad = 2 arc CD, so ∠ cod = 30 degree. In triangle OPD, from cosine theorem, we get PD ^ 2 = OP ^ 2 + od ^ 2 - 2 * OP * od * cos30 = x ^ 2 + 4 - 2 √ 3 * x, so let f (x) = AP + PD = √ (4 + x ^ 2) + √ (x
As shown in the figure, if PA = Pb, ∠ APB = 2 ∠ ACB, AC and Pb intersect at point D, and Pb = 4, PD = 3, then ad · DC=______ ．
Take P as the center of the circle and PA = Pb as the radius to make the circle, extend the intersection circle of BD to m, as shown in the figure: PA = Pb = 4, ∠ APB = 2 ∠ ACB, AC and Pb intersect at point D, PD = 3, let ∠ ACB = θ, then ∠ APB = 2 θ, and ∠ ACB = θ, ｜ C is on the circle. ｜ ad · DC = BD · DM = BD · (PM + PD) = 1 · (4 + 3) = 7, so the answer is 7
As shown in the figure, if PA = Pb, ∠ APB = 2 ∠ C, AC and Pb intersect at point D, and Pb = 4, PD = 3, calculate the value of ad × DC
Take P as the center of the circle, draw a circle through AB, extend BP to D, make PD = 3, and extend to intersect with the circle at m, connect AD and extend to intersect with the circle at C
Then, there is: ∠ C = ∠ AMB (the same as the circumference angle)
Because ∠ APB = 2 ∠ AMB, so ∠ APB = 2 ∠ C
∴AD*DC=MD*DB=1*7=7
As shown in the figure, given the diameter of the circle AB = 6cm, Pb and PC cut semicircle to B and C respectively, PA intersects semicircle to D, and arc AC: arc CB = 1:2, calculate the degree of BPC, PD
1. Let the center of the circle be o, connect OC, ∵ Pb, PC be tangent, ∵ Pb ⊥ AB, PC ⊥ OC ∵ BPC + ∵ BOC = 180 ° ∵ arc AC: arc CB = 1:2 ∵ BC the degree of the arc is 2 / 3 of the semicircle, that is 120 degrees ∵ BPC = 60 ° 2, connect BD, then ∵ ADB = 90 °, connect AC, BC, then the triangle ABC is a right triangle, and ∵ ABC =
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- 14. If the distance between the two ends of line AB and plane α is equal to 1, then the position relationship between line AB and plane α is?
- 15. If there is a line segment A.B on the plane with a length of 5, how many lines are there on the plane with distances of 2 and 3 from point AB?
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