The sum of circumference, diameter and radius of a circle is 46.4cm. What is the circumference and area of the circle?

The sum of circumference, diameter and radius of a circle is 46.4cm. What is the circumference and area of the circle?

3.14*2r+2r+r=(6.28+2+1)r=46.4
r=46.4/(6.28+2+1) = 5cm
Perimeter: 3.14 * 10 = 31.4cm
Area: 3.14 * 25 = 78.5 cm ^ 2

How to find the perimeter and area of a semicircle

The area of a semicircle
=Half the area of a circle of the same radius
=πr²/2
The circumference of a semicircle
=Half of circumference of circle with same radius + diameter of circle
=2πr/2+2r
=πr+2r
=（π+2)r
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The degree of ∠ AOB = ∠ cod = 90 °, BOC: ∠ AOD = 7:11
Please

Let ∠ AOC = a, because ∠ AOB = ∠ cod = 90 °
Then, BOC = 90-a and AOD = 90 + a
Because, BOC: AOD = 7:11
(90-A)/(90+A)=7:11
The solution is a = 20 degree
∠AOC=20°

As shown in the figure, ∠ AOB and ∠ COD are right angles. If ∠ BOC: ∠ AOD = 7:11, calculate the degree of ∠ AOC

∫ AOB and ∠ COD are both right angles, ∫ AOC + ∠ BOC = ∠ BOC + ∠ BOD = 90 °, ∫ AOC = ∠ BOD, ∫ BOC: ∫ AOD = 7:11, ∫ let ∫ BOC = 7x °, ∫ AOD = 11x °, ∫ AOC + ∠ BOD = ∠ AOD - ∠ BOC = 4x °, ∫ AOC = 2x °, ∫ 2x + 7x = 90, the solution is: x = 10, ∫ AOC =

As shown in the figure, given the angle AOC = angle BOC = 100 degrees, and the angle AOB: angle AOD = 2:7, calculate the degree of angle BOC and angle cod

∠ cod = 120 degrees
∠ BOC = 220 degrees
From the known results, we can get: ∠ AOB: (∠ AOB + ∠ BOD) = ∠ AOB: (∠ AOB + 100 °) = 2:7
From this, we can get ∠ AOB = 40 ° so ∠ AOD = 140 ° because it is centered on O, so ∠ cod = 360 ° - 100 ° - 100 ° - 40 ° = 120 ° because ∠ BOC = ∠ cod + ∠ BOD, so ∠ BOC = 220

The ∠ AOC and ∠ BOC complement each other, and the residual angle of ∠ AOC is 10 ° larger than that of ∠ BOC by 3 times

Let the degree of BOC be x and AOC be 3 (90-x) + 10
Because the two corners complement each other, the sum equals 180
x+270-3x+10=180
x=50
Angle AOC = 130

The degree of ∠α and ∠β can be calculated by the degree of ∠α, which is 30 ° larger than 2 times of ∠β

∵∠ α and ∠ β complement each other, and ∵∠ α + β = 180 ° ①, and ∵∠ α is 30 ° larger than 2 times of ∠ β, and ∵∠ α = 2 ∠ β + 30 ° ②, ① and ② are combined. The solution is ∠ α = 50 ° and ∠ β = 130 °. A: the degree of ∠ α is 50 ° and the degree of ∠ β is 130 °

Given vector OA = vector a, vector ob = vector B, and | vector a | = | vector B | = 4, angle AOB = 60 degrees, then | vector a + vector B | =, | vector a-vector B|=

8 and 4 draw a picture, 1 is the middle line of the triangle, 2 is the third side of the triangle

AB is the diameter of circle O, AC is the chord, the line EF is tangent to circle O at point C, and ad ⊥ EF

Prove: connect OC, OA = OC, then ∠ OAC = ∠ OCA;
If EF is tangent to C, OC is perpendicular to CD, and ad is perpendicular to CD
Therefore, OC ∥ ad, which is equal to AC ∥ bad

As shown in the figure, point a is on circle O, circle a intersects circle O at two points F and E, the chord ab of circle O intersects circle a at C, and intersects EF at D. try to explain that AC square = AB times ad
The more complete the answer, the better. I need it urgently!

Because circle a intersects circle O at two points F and E, the chord ab of circle O intersects circle a at C, so AE = AF = AC, so angle AFE = angle AEF, because angle AEF and angle ABF are in the same arc AF, so angle AEF = angle ABF, because angle AFE = angle AEF, so angle AFE = angle ABF, because angle fad = angle BAF, angle AFE = angle ABF, so triangle fad is similar to triangle BAF