The judgment of vertical line and plane in solid geometry of senior high school mathematics

1. If a line is perpendicular to two intersecting lines in the plane, the line is perpendicular to the plane
2. Two parallel lines, one perpendicular to the plane, the other perpendicular to the plane
3. If two surfaces are perpendicular, the line perpendicular to the intersection line in one plane is perpendicular to the other plane
4. Vector method is to prove the perpendicularity of line and line by using vector product being zero, then two vectors are perpendicular, and then method 1 is used to prove line line and line vertical

"Prove the judgment theorem of line and plane parallelism by the method of proof to the contrary"

It is proved that let a ‖ straight line B, a is not in plane α, B is in plane α
Suppose that if a line outside the plane is parallel to a line in the plane, then the line is not necessarily parallel to the plane
If the straight line a is not parallel to the plane α, and because a is not in the plane α, then a intersects with α. Let a ∩ α = F
Make a straight line C ∥ B in plane α through point F,
Because a ‖ B, then a ‖ C
This is in contradiction with a ∩ C = F. therefore, the hypothesis is incorrect and the original proposition is correct

Proof of parallelogram judgment theorem

Parallelogram determination theorem:
1. Two groups of paralleled quadrilateral
2. A quadrilateral whose diagonals are bisected
3. A group of parallelograms whose opposite sides are parallel and equal
3. Two groups of quadrilateral with equal opposite sides

Proving the theorem of parallelogram 2,3

1. Proof: connect AC, ∵ ad = BC, ab = CD, AC = Ca, ∵ ABC ≌ Δ CDA,  ACB = ∠ DAC, ∵ BAC = ≌ ad ≌ BC, ab ∥ CD, ∵ the quadrilateral ABCD is a parallelogram

The judgement theorem of parallelogram

The opposite sides of the two groups were parallel;
The opposite sides of the two groups were equal;
A group of opposite sides are parallel and equal;
4. The diagonals bisect each other;
The diagonal angles of the two groups are equal
The above five conditions can determine that a quadrilateral is a parallelogram and all are parallelogram

All theorems that can prove that a graph is a parallelogram

2. A group of quadrilateral whose opposite sides are parallel and equal is parallelogram. 3. Two groups of quadrilateral whose opposite sides are equal are parallelogram. 4. Two groups of quadrilateral whose opposite sides are parallel are parallelogram. 5. Two groups of quadrilateral whose opposite sides are equal are parallelogram. 6. Two groups of quadrilateral whose opposite sides are parallel and equal are parallel 7. A quadrilateral whose two adjacent corners are complementary is a parallelogram

There are several methods to calculate volume in solid geometry. Thank you

The point m is the midpoint of AB, and M is the midpoint of ab. therefore, am = BM = cm = AC = 2, BC = 2 √ 3, because AB = 2 √ 2, because AB = 2 √ 3, because AB = 2 √ 2, AC = 2 √ 3, because AB = 2 √ 2, AC = 2, BC = 2 √ 3, because AB = 2 √ 2, AC = 2, BC = 2 √ 3, so △ ABC is a right triangle, so the area of △ ABC is 1 / 2 × 2 × 2 √ 2 = 2 √ 2, because AB = 2 √ 2, am = 2 √ 2 because AB = 2 √ 2, am = 2 √ 2, am = 2 √ 2, because AB = 2 √ 2, am = so △ ABM is an isosceles right triangle, In other words, n is the midpoint of AB and Mn = √ 2, because △ ABC is a right triangle, and F is also the midpoint of BC, so FN = 1 / 2Ac = 1 because f is the midpoint of BC and M is the midpoint of AC. therefore, when the right angle △ ABC is extended horizontally, FM ⊥ BC, FM = 1 / 2Ac = 1 (return to the solid) because FM = FN = 1, Mn = √ 2, so ⊥ FMN is isosceles right triangle, so FM ⊥ FN, and because FM ⊥ BC, FM ⊥ plane ABC, So FM is the height of ABCM of triangular pyramid with ABC as the base, so volume = 1 / 3 ×△ ABC area × FM = 2 √ 2 / 3

Volume problem in solid geometry In this case, we find that the volume of BDAC = 1 is vertical to the prism BDAC = 1 Why can we say that ABCD is a square if AC = BD and AC is perpendicular to BD? The title does not say that it is a parallelepiped, only a straight prism, so ABCD does not say it is a parallelogram!

The volume is 3 * 3 * 4, which is 36
The problem solving process is as follows:
Let's first assume that the bottom surface is a square, then the bottom surface is 3 * 3. OK, now we translate one of the diagonal lines, then the bottom surface is isosceles trapezoid, but you will find that the bottom surface is not a square, but an isosceles trapezoid, The extra right triangle is congruent with the missing one, so the base area remains the same
2. Then the bottom area is 3 * 3
Then the volume can be calculated
I'm from Jiangsu, I don't know if you are satisfied with the answer

Solid geometry volume problem of cylinder and cone It is known that E and F are the mid points of edge AB, CD of regular tetrahedron ABCD whose edge length is a. rotate triangle AEF around AF and calculate the volume of the rotator I'll make the root of 18 3 But the answer is 36 th root 3

It is easy to know Fe ⊥ AB from the three perpendicular theorem, so Δ AEF is RT triangle, and AE = A / 2, AF = a √ 3 / 2, and then EF = a √ 2 / 2 from Pythagorean theorem
For RT Δ AEF, the square of the distance from e to AF (that is, the square of the height on the slope) is a 2 / 6
Therefore, the volume obtained is 1 / 3 × (a 2 / 6) π × (a √ 3 / 2) = (√ 3) a 3 π / 36

Surface area, side area, total area and volume formula of all graphics from primary school to senior high school

Rectangle: S = ab {rectangle area = length × width}
Square: S = a ^ 2 {square area = side length × side length}
Parallelogram: S = ab {parallelogram area = base × height}
Triangle: S = ab △ 2 {triangle area = bottom × height △ 2}
Trapezoid: S = (a + b) × h △ 2 {trapezoid area = (upper bottom + bottom) × height △ 2}
Circle (circle): S = Πr ^ 2 {area of circle (circle) = circumference × radius × radius}
Circle (circular outer ring): S = Πr ^ 2 - Πr ^ 2 {area of circle (outer ring) = circumference × outer ring radius × outer ring radius - circumference ratio × inner ring radius × inner ring radius}
Circle (sector): S = Πr ^ 2 × n / 360 {area of circle (sector) = circumference × radius × radius × fan angle / 360}
Cuboid surface area: S = 2 (AB + AC + BC) {cuboid surface area = (length × width + length × height + width × height) × 2}
Cube surface area: S = 6A ^ 2 {cube surface area = edge length × edge length × 6}
Surface area of sphere (positive ball): S = 4 Π R ^ 2 {surface area of sphere (positive ball) = circumference × radius × radius × 4}
The ellipse s = π (PI) × a × B (where a and B are the length of the long half axis and the short half axis of the ellipse respectively)

The judgment of vertical line and plane in solid geometry of senior high school mathematics