Known arc length 8.8 chord length 6.6 radius

Known arc length 8.8 chord length 6.6 radius

Let radius = R, arc length = s, chord length = L. there are: r = s / 2arcsin (L / 2R)

Given the chord length and chord height, it is better to have a formula
Group 1: chord length 3.9m 5m
Group 2: chord length 3.7m, chord height 0.89m
How to find the arcuate area?
It's better to have a formula

The area of arc AB less than semicircle = the area of sector OAB - △ OAB
The area of arc AB larger than semicircle = the area of sector OAB + △ OAB
The area of fan-shaped OAB = ∠ the degree of AOB * π R ^ 2 / 360, where o is the center of the circle, AB is the chord, π is the circumference, and R is the radius
1. Let o be OC ⊥ AB over C, then AC = CB = 1.95, OC = r-1.5,
OA^2=OC^2+AC^2,
∴r^2=(r-1.5)^2+1.95^2,
3r=6.0525,
r=2.0175,r-1.5=0.5175,
sinAOC=AC/OA=0.96654275,
∠AOC=75.13718385°,
∠AOB=150.2743677°,
Arcuate area = 150.2743677 * π * 2.0175 ^ 2 / 360-3.9 * 0.5175/2
=4.328638996(m^2).
2. Leave it to you to practice

Urgent bow area formula, known chord length C, height h, help calculate the area, to formula, results
1. Chord length 8.5, height 0.075
2. Chord length 8.5, height 0.095
3. Chord length 11.5, height 0.075
4. Chord length 8.5, height 0.095

Arcuate area f = 1 / 2 [R L-C [R-H]]; r = (C square + 4H Square) △ 8h; L = 0.017453ar; a = 57.296l/r
Where R is the radius of the circle, C is the chord length, h is the height, l is the arc length, and a is the radian

If m, n ∈ {x | x = A2 × 102 + A1 × 10 + A0}, where AI ∈ {1, 2, 3, 4, 5, 6, 7}, I = 0, 1, 2, and M + n = 636, then the number of different points on the plane is ()
A. 60 B. 70 C. 90 D. 120

Let a = {x | x = A0 + A1 · 10 + A2 · 100}, the number of different points on the coordinate plane (x, y) is equivalent to the number of solutions of X + y = 636 in a, which can be investigated according to the decimal bit. First, look at the individual bit, A0 + A0 = 6, there are five possibilities. Then look ahead: a1 + A1 = 3 and A2 + A2 = 6, there are 2 × 5 = 10 possibilities

If (1 + MX) 6 = A0 + a1x + a2x2 + +A6x6, and a1 + A2 + +A6 = 63, then the value of real number m is______ ．

Let x = 0, we can get A0 = 1, let x = 1, we can get (1 + m) 6 = A0 + A1 + A2 + +a6，∴a1+a2+… +a6=（1+m）6-1∵a1+a2+… +A6 = 63, 1 + m 6-1 = 63, M = 1 or - 3, so the answer is: 1 or - 3

What formula is n pie r square / 360?

Sector area formula

How to deduce the quadratic power / 360 of sector formula n π r?

1. The area of the garden is π r quadratic
2. 1 circle angle = 360 degrees
3. The degree of the central angle is a fraction of the circumference, and the area of the sector is a fraction of the area of the garden
So: sector formula = the second power of π R n / 360

How can two formulas of cone side area formula be equal (n / 360 × π × R2 = 1 / 2lr), please explain each formula in detail and explain how to be equal

There are two formulas for the side area of a cone
S = 1 / 2RL
S = π RL
They are all right, but in different ways
To find the side area of a cone, we must first deform the cone
Suppose you cut a generatrix along the cone, and then expand it to get a sector
There are two methods to calculate the sector area, and the results have the above two different expressions
Expression 1
Using the integral principle
Suppose that the sector is made up of N isosceles triangles, which are small enough to have the base length = R / N (R is the circumference of the ground circle of the cone, i.e. the arc length of the sector), and the height = the side length L (L is the radius of the sector, i.e. the generatrix of the cone)
Then the sector area
S = n (number of triangles) x s (unit area of isosceles triangle)
= n X (1/2 X R/n X L)
= 1/2RL
Expression 2
Use arc length
Sector area / total area of circle = arc length / circumference
Sector area
S = total area of circle (circle of sector) x (arc length / circumference)
=Total area of circle x (ground perimeter of cone / circle perimeter of sector)
=π L2 (L is the bus length) x (2 π R / 2 π L)
= πLR

S = n π R ^ 2 △ 360, then r =? Formula

R = (?) 10004; 360s / n

Calculation formula of cone area
Use letters and make notes

There are two formulas
1. S = one-third × π (pie, I'm afraid you can't understand it, that's 3.14) R & sup2; × H (one-third bottom area multiplied by height) This primary school should learn it
2. S = π × R × L (L is the generatrix, the generatrix is the line on the side of the cone)
S=π×r×l+πr²…… This junior high school
You can click on this picture,